A $8,000 principal is invested in two accounts, one earning 1% interest and another earning 6% interest. If the total interest for the year is $405, then how much is invested in each account?

The dimensions (length and width) of the poster are 6 inches and 2 inches

Represent the length with x and the width with y

From the question, we have the following parameters

x = 5 + 0.5y

Area, A = 12

The area of a rectangle is represented as:

A = xy

Substitute the known values in the above equation

(5 + 0.5y) * y = 12

Expand the bracket

5y + 0.5y^2 = 12

Multiply through by 2

10y + y^2 = 24

Rewrite as:

y^2 + 10y - 24 = 0

Expand

y^2 + 12y - 2y - 24 = 0

Factorize the expression

(y - 2)(y + 12) = 0

Solve for y

y = 2 or y = -12

The dimension cannot be negative.

So, we have

y = 2

Substitute y = 2 in x = 5 + 0.5y

x = 5 + 0.5 * 2

Evaluate

x = 6

Hence, the dimensions (length and width) of the poster are 6 inches and 2 inches

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Answer:

1.

Step-by-step explanation:

10C0 = 1.

The Area of the shaded region of the circle = 60.75π m²

The Length of arc ADB = ∅/360 × 2πr = 13.5π m.

The area of the shaded region in the circle = area of the sector of a circle = ∅/360 × πr², where r is the radius of the circle and the central angle is ∅.

The length of the arc on a circle = ∅/360 × 2πr, where r is the radius of the circle and the central angle is ∅.

Given the following:

Central angle (∅) = 360 - 90 = 270°

Radius (r) = 9 m.

Area of the shaded region of the circle = ∅/360 × πr² = 270/360 × π(9²)

Area of the shaded region of the circle = 270/360 × π81

Area of the shaded region of the circle = 60.75π m²

Length of arc ADB = ∅/360 × 2πr = 270/360 × 2π(9)

Length of arc ADB = ∅/360 × 2πr = 270/360 × 18π

Length of arc ADB = ∅/360 × 2πr = 13.5π m.

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Answer:last option -5± √17

2

Step-by-step explanation:

Quadratic equations represent any equation that can be rearranged in standard forma (ax² + bx + c( =0(a, b & c) are known.

Quadratic equations can always be solved using the quadratic formula, but sometimes factoring or isolating the variable is also posible.

In the square equation ax² + bx + c = 0

[tex]\boldsymbol{\sf{x=\dfrac{-b\pm\sqrt{\Delta} }{2a} \ ,\Delta=b^{2}-4ac } }[/tex]

Let's calculate the discriminant of the quadratic equation:

Since the discriminant is greater than zero, then the quadratic equation has two real roots.

[tex]\boldsymbol{\sf{x_{1}=\dfrac{-b-\sqrt{\Delta} }{ 2\cdot a}=\dfrac{-5-\sqrt{17} }{2\cdot1} }}[/tex]

[tex]\boldsymbol{\sf{x_{2}=\dfrac{-b+\sqrt{\Delta} }{ 2\cdot a}=\dfrac{-5+\sqrt{17} }{2\cdot1} }}[/tex]

Solution:

[tex]\boldsymbol{\sf{x=\dfrac{-5\pm\sqrt{17} }{2} }}[/tex]

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Answer:12002/9987

Step-by-step explanation:

The number 0.22 represents the population proportion and the correct way to represent it is p = 0.22.

Statistics is a mathematical tool defined as the study of collecting data, analysis, understanding, representation, and organization. Statistics is described as the procedure of collecting data, classifying it, displaying that in a way that makes it easy to understand, and analyzing it even further.

It is given that:

It is given that:

The Department of Motor Vehicles reports that the proportion of all vehicles registered in California that are imported is 0.22.

The number 0.22 represent the population proportion.

The population proportion can be represented by the letter p:

p = 0.22

Thus, the number 0.22 represents the population proportion, and the correct way to represent it is p = 0.22.

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The equation of the line that passes through the given points is

y = 1/3x + 5.

The formula for the equation of a line passing through two points (x1, y1) and (x2, y2) is

[tex](y-y1) = \frac{(y2-y1)}{(x2-x1)} (x-x1)[/tex]

Where the fraction (y2 - y1)/(x2 - x1) is the slope of the line denoted by 'm'.

It is given that, a line passes through the points (-3,4) and (6,7).

So, the equation of the line is

[tex](y-y1) = \frac{(y2-y1)}{(x2-x1)} (x-x1)[/tex]

On substituting x1 = -3, y1 = 4, x2 = 6, and y2 = 7

(y - 4) = [(7 - 4)/(6 + 3)](x + 3)

⇒ (y - 4) = (3/9)(x + 3)

⇒ (y - 4) = 1/3(x + 3)

⇒ y- 4 = 1/3x + 1

⇒ y = 1/3x + 1 + 4

∴ y = 1/3x + 5

Thus, the equation of the line passing through the points (-3,4) and (6,7) is y = 1/3x + 5. Where the slope m = 1/3.

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The value of cos(L) in the triangle is Five-thirteenths

Right triangles are triangles whose one of its angle has a measure of 90 degrees

The value of a cosine function is calculated as:

cos(L) = Adjacent/Hypotenuse

The hypotenuse is calculated as

Hypotenuse^2 = Opposite^2 + Adjacent^2

So, we have:

Hypotenuse^2 = 12^2 + 5^2

Evaluate

Hypotenuse^2 = 169

Take the square root of both sides

Hypotenuse = 13

So, we have

Adjacent = 5

Hypotenuse = 13

Recall that

cos(L) = Adjacent/Hypotenuse

This gives

cos(L) = 5/13

Hence, the value of cos(L) in the triangle is Five-thirteenths

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Answer:

40 + 10 + 30 = 50 + 30 = 80

Step-by-step explanation:

40 + 10 + 30 = 50 + 30 = 80

Answer:

10(4+3+1)

Step-by-step explanation:

Expand 40+30+10 by the distributive property:: 10(4+3+1)

Answer:

3.2

Step-by-step explanation:

4 x 8 / 8 + 2

32/10

=3.2

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The angles and length of the triangle area as follows:

The sum of angles in a triangle is 180 degree.

Therefore,

∠DBC = 180 - 29 - 90 = 61°

Hence,

∠A = 180 - 29 - 61 - 55 = 35°

∠CBD = 61°

Using trigonometric ratios,

sin 55 = opposite / hypotenuse

sin 55 = AD / 40

AD = 40 sin 55

AD = 32.7660817716

AD = 32.77 units

cos 55 = adjacent / hypotenuse

cos 55 = BD / 40

BD = 40 cos 55

BD = 22.943057454

BD = 22.94 units

sin 29 = 22.94 / BC

BC = 22.94 sin 29

BC = 11.1215326885

BC = 11.12 units

cos 29 = CD / 11.12

CD = 11.12 cos 29

CD = 9.72577114339

CD = 9.73 units

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Answer:

12km/h

Step-by-step explanation:

s=d/t

900m is 0.9km and 4min and 30sec is 0.075h

0.9/0.075=12 which gives us 12km/h

Answer:

Step-by-step explanation:

4 min 30 seconds = 270 seconds

900m = 270 seconds

100m = 30 seconds

1000m= 300 seconds (5min)

12,000m=3600seconds (60min)

Answer = 12km/h

Answer:

Step-by-step explanation:

The sum of interior angles of a quadrilateral is 360°.

If we consider the measure of one of the unknown angles to be [tex]x[/tex]°, we can set up the following equation:

[tex]x + x + 75^\circ + 117^\circ = 360^\circ[/tex]

Now we can solve for [tex]x[/tex]:

⇒ [tex]2x + 192^\circ = 360^\circ[/tex]

⇒ [tex]2x = 360^\circ - 192^\circ[/tex]        [subtracting 192° from both sides]

⇒ [tex]2x = 168^\circ[/tex]

⇒ [tex]x = \frac{168^\circ}{2}[/tex]                      [dividing both sides by 2]

⇒ [tex]x = \bf84^\circ[/tex]

Therefore, the other two angles each have a measure of 84°.

Step-by-step explanation:

[tex]y = \sqrt{4x + 6} [/tex]

[tex] {y}^{2} = 4x + 6[/tex]

[tex] {y}^{2} - 6 = 4x[/tex]

[tex] \frac{1}{4} ( {y}^{2} - 6) = x[/tex]

Swap x and y

[tex] \frac{ {x}^{2} - 6 }{4} = y[/tex]

Answer:

Perimeter o the rectangle = 16x - 2

Step-by-step explanation:

Perimeter of a rectangle = 2(long+ wide)

Then:

Perimeter = 2((6x-4)+(2x+3))

Perimeter = 2(6x+ 2x + 3 - 4)

Perimeter = 2(8x - 1)

Perimeter = 2*8x + 2*-1

Perimeter = 16x - 2

y=2

-5(2)²-3(2)-2

-5(4)-3(2)-2

-20-6-2

=12

Answer:

x=3/2 y=1/2

Step-by-step explanation:

1. Solving linear equations:

(a) x = 4.

(b) r = 2/3.

2. Proportion:

(a) When x = 40, y = 100.

(b) When x = 40, y = 4.

3. System of equations:

(a) r = 8, s = 3.

(b) p = 5, q = -2.

4. Graphing lines:

(a) The slope of the line through (3, 4) and (-1, 3) is 1/4.

(b) The slope of the line 3x - 4y = 7 is 3/4.

(c) The slope-intercept form of the line through (5, -2) and (-1, 6) is y = (-4/3)x + 14/3.

5. Introductory quadratics:

(a) The solutions to the equation 4x² = 81 are -9/2, and 9/2.

(b) The solutions to the equation x² + 8x + 12 are -6, and -2.

(c) The solutions to the equation x² - 3x - 88 are -8, and 11.

6. The weight of an orange is 1 unit, the weight of an apple is 5 units, and the weight of a banana is 2 units.

7. x = 3.

1. Solving linear equations:

(a) 3x - 7 = 9 - x,

or, 3x + x = 9 + 7,

or, 4x = 16,

or, x = 16/4 = 4.

Thus, x = 4.

(b) (7 - 2r)/3 = 4r,

or, 7 - 2r = 3*4r = 12r,

or, -2r - 12r = -7,

or, -14r = -7,

or, r = -7/-14 = 1/2.

Thus, r = 1/2.

2. Proportion:

(a) x and y directly proportion, means x/y = constant.

When x = 8, y = 20.

Thus, x/y = constant, or, 8/20 = constant, or, constant = 0.4.

When x = 40,

x/y = 0.4,

or, y = x/0.4 = 40/0.4 = 100.

Thus, when x = 40, y = 100.

(b) x and y indirectly proportion, means xy = constant.

When x = 8, y = 20.

Thus, xy = constant, or, 8*20 = constant, or, constant = 160.

When x = 40,

xy = 160,

or, 40y = 160,

or, y = 160/40 = 4.

Thus, when x = 40, y = 4.

3. System of equations:

(a) r - s = 5 ...(i)

3r - 5s = 9 ... (ii)

3*(i) - (ii) gives:

3r - 3s = 15

3r - 5s = 9

(-) (+)    (-)

_________

2s = 6,

or, s = 3.

Substituting in (i), we get

r - s = 5,

or, r - 3 = 5,

or, r = 8.

Thus, r = 8, s = 3.

(b) 3p + 7q = 1 ...(i).

5p = 14q + 53,

or, 5p - 14q = 53 ...(ii).

2*(i) + (ii) gives:

6p + 14q = 2

5p - 14q = 53

____________

11p = 55,

or, p = 5.

Substituting in (i), we get:

3p + 7q = 1,

or, 3*5 + 7q = 1,

or, 7q = 1 - 15 = -14,

or, q = -14/7 = -2.

Thus, p = 5, q = -2.

4. Graphing lines:

(a) Slope of the line through (3, 4) and (-1, 3) is,

m = (4 - 3)/(3 - (-1)),

or, m = 1/4.

Thus, the slope of the line through (3, 4) and (-1, 3) is 1/4.

(b) The graph given: 3x - 4y = 7.

Representing in the slope-intercept form, y = mx + b, gives:

3x - 4y = 7,

or, 4y = 3x - 7,

or, y = (3/4)x + (-7/4).

Thus, the slope of the line 3x - 4y = 7 is 3/4.

(c) Slope of the line through (5, -2) and (-1, 6) is,

m = (6 - (-2))/(-1 - 5),

or, m = 8/(-6) = -4/3.

Substituting m = -4/3 in the slope-intercept form, y = mx + b, gives:

y = (-4/3)x + b.

Substituting y = 6, and x = -1 gives:

6 = (-4/3)(-1) + b,

or, b = 6 - 4/3 = 14/3.

Thus, the slope-intercept form of the line through (5, -2) and (-1, 6) is y = (-4/3)x + 14/3.

5. Introductory quadratics:

(a) 4x² = 81,

or, 4x² - 81 = 0,

or (2x)² - 9² = 0,

or, (2x + 9)(2x - 9) = 0.

Either, 2x + 9 = 0 ⇒ x = -9/2,

or, 2x - 9 = 0 ⇒ x = 9/2.

Thus, the solutions to the equation 4x² = 81 are -9/2, and 9/2.

(b) x² + 8x + 12 = 0,

or, x² + 2x + 6x + 12 = 0,

or, x(x + 2) + 6(x + 2) = 0,

or, (x + 6)(x + 2) = 0.

Either, x + 6 = 0 ⇒ x = -6,

or, x + 2 = 0 ⇒ x = -2.

Thus, the solutions to the equation x² + 8x + 12 are -6, and -2.

(c) x² - 3x - 88 = 0,

or, x² - 11x + 8x - 88 = 0,

or, x(x - 11) + 8(x - 11) = 0,

or, (x + 8)(x - 11) = 0.

Either, x + 8 = 0 ⇒ x = -8,

or, x - 11 = 0 ⇒ x = 11.

Thus, the solutions to the equation x² - 3x - 88 are -8, and 11.

6. We assume the weight of one orange, one apple, and one banana to be x, y, and z units respectively.

Thus, we have:

3x + 2y + z = 15 ... (i)

5x + 7y + 2z = 44 ... (ii)

x + 3y + 5z = 26 ... (iii)

2(i) - (ii) gives:

6x + 4y + 2z = 30

5x + 7y + 2z = 44

(-)  (-)  (-)        (-)

______________

x - 3y = -14 ... (iv)

5(i) - (iii) gives:

15x + 10y + 5z = 75

x + 3y + 5z = 26

(-) (-) (-)          (-)

________________

14x + 7y = 49 ... (v)

14(iv) - (v) gives:

14x - 42y = -196

14x + 7y = 49

(-)    (-)     (-)

_____________

-49y = -245,

or, y = 5.

Substituting in (v), we get:

14x + 7y = 49,

or, 14x + 35 = 49,

or, x = 14/14 = 1.

Substituting x = 1 and y = 5 in (i), we get:

3x + 2y + z = 15,

or, 3 + 10 + z = 15,

or, z = 2.

Thus, the weight of an orange is 1 unit, the weight of an apple is 5 units, and the weight of a banana is 2 units.

7. 3/(1 - (2/x)) = 3x,

or, 3/((x - 2)/x) = 3x,

or, 3x/(x - 2) = 3x,

or, 3x = 3x(x - 2),

or, 3x = 3x² - 6x,

or, 3x² - 9x = 0,

or, 3x(x - 3) = 0.

Either, 3x = 0 ⇒ x = 0,

or, x - 3 = 0 ⇒ x = 3.

Since we had a term 2/x, x cannot be 0.

Thus, x = 3.

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Answer:

a=4.40

Step-by-step explanation:

To find value of a use pythagoras theorem:

c^2=b^2+a^2

Rearrange the equation:

a^2=c^2-b^2

Substitute the values:

a^2=(4.58)^2-(1.26)^2

After calculation:

a=4.40

If we compare the given values then we can find that the database B is more likely to have smaller standard deviation.

Given that the values in database A are 0 from 100 and has mean of 50 and Database B has entries from 49 to 51 and also has mean of 50.

We are required to find the database whose standard deviation is lower.

Standard deviation measures the variation of values. It is calculated after finding mean. The square of a standard deviation is known as variance.

Database A has values from 0 to 100 and has mean of 50. Because the values are somewhat very larger than 50 and in database B has values from 49 to 51,there are more chances that the standard deviation of database B will have smaller value than from standard deviation of database A.

Hence if we compare the given values then we can find that the database B is more likely to have smaller standard deviation.

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The two ordered pairs that the mid point is (4,-10) are (4,-20),(4,0) & (4,0),(4,-20).

Given the coordinates of mid point be (4,-10).

We are required to find the ordered pairs that the mid point is (4,-20).

Coordinates show positions of points or something else on a surface.

There are various combinations whose mid point is (4,-10).

First are (4,-20),(4,0).

Mid point =[(4+4)/2,(-20+0)/2]

=(4,-10)

Second are (4,0) , (4,-20)

Mid point=[(4+4)/2,(0-20)/2]

=(4,-10)

Third are (8,-20),(0,0)

Mid point=[(8+0)/2,(-20+0)/2]

=(4,-10)

Fourth are (0,-20),(8,0)

Mid point =[(0+8)/2,(-10+0)/2]

=(4,-10)

Hence the ordered pairs that the mid point is (4,-10) are (4,-20),(4,0) & (4,0),(4,-20)& (8,-20),(0,0)&(0,-20),(8,0)&(0,0),(8,-20).etc.

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A)  Rotate by 90° counterclockwise.

B) Reflection transformation about the line y = 5.

C) Parallel Lines are C₂D₂ and A₂B₂; EF and E₁F₁.

D) Perpendicular lines are;  D₁F₁ and  E₁F₁; DF and EF; DC and AB.

E) Line segments with slope of 0 are; AB, C₂D₂, A₂B₂, EF and E₁F₁

F) Line segments on the boat that have an undefined slope are; Lines A₁B₁, DF and DC.

G) Slope of Line ED = 1

A) The blue segment CD is seen on the graph as a perpendicular line with 2 units while the line segment C₂D₂ is seen as a horizontal line. Thus, to match CD with C₂D₂, we will rotate by 90° counterclockwise.

B) The transformations that would be used on the blue triangle to get it to match with the red triangle is a reflection transformation about the line y = 5.

C) The line segments that are parallel to each other are; C₂D₂ and A₂B₂; EF and E₁F₁.

D) The line segments that are perpendicular are;  D₁F₁ and  E₁F₁; DF and EF; DC and AB.

E) Horizontal lines that are parallel to the x-axis have zero slope. Thus, AB, C₂D₂, A₂B₂, EF and E₁F₁ all have zero slopes.

F) Undefined slope is the slope of a vertical line. Thus, Lines A₁B₁, DF and DC have undefined slopes.

G) Slope of ED = (5 - 4)/(-10 - (-11))

Slope of ED = 1/1

Slope = 1

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1529.3 square unit is the area of the cheetah area at a zoo is designed in a triangular fashion, surrounded on all three sides by sidewalks, given that the property has 67 feet of frontage on one sidewalk, and 48 feet of frontage on another; these two sidewalks intersect at a 72° angle. This can obtained by using the formula of Area of a Triangle with 2 Sides and Included Angle.

Area of a triangle is obtained using the formula of Area of a Triangle with 2 Sides and Included Angle.

If in a triangle ΔABC,

Here it is given that,

67 feet and 48 feet are the sides of the triangular space and angle 72° is the included angle.

By using the formula of Area of a Triangle with 2 Sides and Included Angle,

Area of the cheetah area = 1/2 × bc × sin(A)

Area of the cheetah area = 1/2 × (67)(48) × sin(72°)

sin(72°) = 0.951056516

Area of the cheetah area = 1608 × 0.951056516

Area of the cheetah area = 1529.29888 ≈ 1529.3 square unit

Hence 1529.3 square unit is the area of the cheetah area at a zoo is designed in a triangular fashion, surrounded on all three sides by sidewalks, given that the property has 67 feet of frontage on one sidewalk, and 48 feet of frontage on another; these two sidewalks intersect at a 72° angle.

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The minimum value of z is -38

The optimization equation is given as

z=−3x+5y

The constraints are given as:

x+y≥−2

3x−y≤2

x−y≥−4

Next, we plot the constraints on a graph and determine the points of intersections

See attachment for the graph

From the attached graph, the points of intersections are

(-9, -13) and (-10, -12)

So, we have:

(-9, -13)

(-10, -12)

Substitute these values in the objective function

z=−3x+5y

This gives

z= −3 * -9 +5 * -13 = -38

z= −3 * -10 +5 * -12 = -30

-38 is less than -30

Hence, the minimum value of z is -38

So, the complete parameters are:

Optimization Equation:

z=−3x+5y

Constraints:

x+y≥−2

3x−y≤2

x−y≥−4

Vertices of the feasible region

(0, -2)

(-3, 1)

(3, 7)

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Answer:

(1,2)

Step-by-step explanation:

Substitution:

x + 2y = 5  Solve for x

x = -2y + 5  Substitute -2y + 5 in for x in the second equation

4x - 12y = -20

4(-2y + 5) - 12y = -20  Distribute the 4

-8y + 20 - 12 y = -20  Combine the y term

-20y + 20 = -20  Subtract 20 from both sides

-20y = -40  Divide both sides by -20

y = 2

Plug y into either of the 2 original equations to get x.

x + 2y = 5

x + 2(2) = 5

x + 4 = 5

x = 1

The answer is (2,1).

Elimination:

x + 2y = 5      4x - 12y = -20.  We want to eliminate with the x or the y.  I am going to eliminate the x's that means that I have to multiply the first equation all the way through by -4

(-4)(x + 2y) = (5) (-4)  That makes the equivalent expression

-4x - 8y = -20  I will add that to 4x - 12y = -20

4x - 12y = -20

0x -20y = -40

-20y = -40

y = 2.  Plug 2 into either the 2 original equation to find x.  This time I will select the second original equation to find x.

4x -12y = -20

4x - 12(2) = -20

4x - 24 = -20

4x = 4

x = 1

WhenWhen we divide 8x⁴ - 6x³ + 2 by 2x² + 3 The result obtained is 4x² - 3x - 6 remainder 9x + 20

Quotient is the result obtained when division operation is carried out.

For example when 6 is divided by 2, the result obtained is 3. Thus, the quotient is 3

To divide 8x⁴ - 6x³ + 2 by 2x² + 3, we shall apply the long division method. This is illustrated below:

                          4x² - 3x - 6        

             2x² + 3 | 8x⁴ - 6x³ + 2

                         -(8x⁴ + 12x²)

                         -6x³ - 12x² + 2

                        -(-6x³ - 9x)

                        -12x² + 9x + 2

                         -(-12x² - 18)

                          9x + 20            

Thus, the result obtained when we divide 8x⁴ - 6x³ + 2 by 2x² + 3 is 4x² - 3x - 6 remainder 9x + 20

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Due to length restrictions, we kindly invite to check the explanation herein for further details of the hyperbola.

Herein we have an hyperbola whose axis of symmetry is parallel to the y-axis and the major semiaxis length is in the y-direction. By analytical geometry, we know that eccentricities of hyperbolae are greater than 1.

a) The formula for eccentricity is:

e = √(a² + b²) / a      (1)

Where:

If we know that a = 4 and b = 3, then the eccentricity of the hyperbola is:

e = √(4² + 3²) / 4

e = 5 / 4

b) The coordinates of the two vertices of the hyperbola are:

V(x, y) = (h, k ± a)      (2)

Where (h, k) are the coordinates of the center of the hyperbola.

V₁ (x, y) = (0, 4), V₂ (x, y) = (0, - 4)

The coordinates of the foci of the hyperbola are:

F(x, y) = (h, k ± c), where c = √(a² + b²).     (3)

c = √(4² + 3²)

c = 5

F₁ (x, y) = (0, 5), F₂ (x, y) = (0, - 5)

The equations of the asymptotes of the hyperbola are:

y = ± (a / b) · x

y = ± (4 / 3) · x      (4)

And the equations of the directrices of the hyperbola are:

y = k ± (2 · a - c)

y = 0 ± (8 - 5)

y = ± 3     (5)

The graph is presented in the image attached below.

c) The parametric equations for the hyperbola are the following formulae:

y = ± a · cosh t   →   y = ± 4 · cosh t      (6)

x = b · sinh t   →   x = 3 · sinh t     (7)

d) First, we determine the slopes of the two tangent lines by implicit differentiation:

m = (16 · x) / (9 · y)

m = (16 · 2.3) / [9 · (± 4.807)]

m = ± 0.851

Second, we find the intercept of each tangent line:

(x, y) = (2, 4.807)

b = 4.807 - 0.851 · 2

b = 3.105

y = 0.851 · x + 3.105      (8)

(x, y) = (2, - 4.807)

b = - 4.807 - (- 0.851) · 2

b = - 3.105

y = - 0.851 · x - 3.105      (9)

e) The definite integral of the arc length of the hyperbola is presented below:

[tex]s = \int\limits^{2}_{1} {\sqrt{\left(\frac{dx}{dt} \right)^{2}+\left(\frac{dy}{dt} \right)^{2}}} \, dt[/tex]

If we know that dx / dt = a² · sinh² t and dy / dt = b² · cosh² t, then the definite integral for the arc length is:

[tex]s = \int\limits^2_1 {\sqrt{a^{2}\cdot \sinh ^{2}t +b^{2}\cdot \cosh^{2}t}} \, dt[/tex]      (10)

f) We apply the following substitutions on (1): x = r · cos θ, y = r · sin θ. Then, we have the polar form by algebraic handling:

r(θ) = (a · b) / (b² · sin² θ - a² · cos² θ)     (11)

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1a. 98 cm ^2

1b. 76. 98 cm^2

2a. 42cm

2b. 7. 19 cm

3. a + b/2 (h)

4. 24 cm ^2

5. πr + 2r

6. 13. 2m

7. 45cm^2

8. 252 cm ^ 2

9. 450 cm^ 2

1a. The shape given is a rectangle

The formula for area of a rectangle is given as;

Area = length × width

Area = 7 × 14

Area = 98 cm ^2

1b The shape given is a semi circle

The formula for area of a semicircle is given as;

Area = 1/2 π r^2

radius = diameter/2 = 14/2 = 7cm

Area = 1/2 × 3.142 × 7 × 7

Area = 76. 98 cm^2

2a. The shape given is a rectangle

The formula for perimeter of a rectangle is given as;

Perimeter = 2 ( length + width)

Perimeter = 2 ( 14 + 7) = 2( 21)

Perimeter = 42cm

2b. The shape is a semicircle

Perimeter = π r + 2r

r= 1.4cm; diameter divided by 2

Perimeter = 3. 142(1.4) + 2(1.4)

Perimeter = 7. 19 cm

3. The formula for area of a trapezium is given as

Area = a + b/2 (h)

4. The area of the trapezium is given as;

Area = 9 + 7/2 (3)

Area = 16/2 (3)

Area = 8 × 3

Area = 24 cm ^2

5. Area of semicircle = 1/2 πr^2

Perimeter of a semicircle = πr + 2r

6. From the information given, we have the following

Area = 480 m^2

a = 20m

b = unknown

h = 13. 2m

Area = a+b/2 (h)

Substitute the values

480 = 20+b/2 (13. 2)

480 = 10+ b (13. 2)

480/13. 2 = 10 + b

10+ b = 480/ 13. 2

10 + b = 36. 36

b = 36.36 - 10

b = 26. 36m

7. The formula for area of a rhombus is given as

Area = p × q/2

Where p and q are the diagonals

Area = 7. 5 × 12/2

Area = 90/2

Area = 45cm^2

8. The formula for area for a quadrilateral is given as;

Area of quadrilateral = (½) × diagonal length × sum of the length of the perpendiculars

sum of the length = 13 + 8 = 21cm

Diagonal A= 24cm

Area = 1/2 × 24 × 21

Area = 252 cm ^ 2

9. Area of a pentagonal park = 1/2 × sum of parallel sides × height

Sum of parallel sides = 15 + 15 = 30 cm

height = 30cm

Area = 1/2 × 30 × 30

Area = 450 cm^ 2

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Answer: I believe the answer is 25.

Step-by-step explanation:

You convert 1/5 to a decimal.

1/5 = 0.2

then you divide

180 divided by 7.2 = 25

answer is 25.

The number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

Note that the formula for combination is given as;

Combination = [tex]\frac{n!}{r!(n-r)!}[/tex]

From the information we have that;

There are 28 questions in the pool

The test should have a total of 10 questions;

6 difficult , 10 average and 12 easy questions

We are asked to determine the combination of;

3 difficult questions

4 average questions

3 easy questions

6C3 = [tex]\frac{6!}{3!(6-3)!}[/tex]

6C3 = [tex]\frac{720}{36}[/tex]

6C3 = 20

10C4 = [tex]\frac{10!}{4!(10-4)!}[/tex]

10C4 = [tex]\frac{3628800}{17280}[/tex]

10C4 = 210

12C4 = [tex]\frac{12!}{4!(12-4)!}[/tex]

12C4 = 220

The number of ways of constructing the questions is

= 20 × 210 × 220

= 924, 000 ways

Thus, the number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

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