Question 8
What is wrong with the following equation: 8 + (6/2) = 17-5
The loft

Answer:

- [tex]\sqrt{2}[/tex]

Step-by-step explanation:

cos90° - 2sin45° + 2tan180°

= 0 - ( 2 × [tex]\frac{\sqrt{2} }{2}[/tex] ) + 2(0)

= 0 - [tex]\sqrt{2}[/tex] + 0

= - [tex]\sqrt{2}[/tex]

The correct option regarding whether the relation is a function is:

B. The relation is a function. Each input value is paired with one output value.

A relation represent a function if each value of the input is paired with one value of the output.

In this problem, when the input - output mappings are given by:

{(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)}.

Which means that yes, each input value is paired with one output value, hence the relation is a function and option B is correct.

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Given that the pieces of steel rods comes in bundles, the mechanist will require 13 bundles of steel rods to get the 98 pieces of steel rod he needs.

Given the data in the question;

To determine the bundle of steel required, let y represent the bundle.

Since;

1 bundle = 8 piece

y bundle = 98 piece

We cross multiply

y bundle × 8 piece = 1 bundle × 98 piece

y = ( 1 bundle × 98 piece ) / ( bundle ×  8 piece  )

y = 98 pieces / 8 piece

y = 12.25 ≈ 13

Given that the pieces of steel rods comes in bundles, the mechanist will require 13 bundles of steel rods to get the 98 pieces of steel rod he needs.

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Check the picture below.

notice, the dividend and divisor must be in descending order, and when one of the variables is "missing", is really not missing, it simply has a coefficient of 0.

Here is the solution process:

                     3 x² -  2 x  -   6                      

x² + 0x + 2 | 3 x⁴ -  2 x³ +  0  x² -     x -   4

                    3  x⁴ + 0 x³ +  6  x²                

                             - 2 x³  -  6  x² -      x -   4

                             - 2 x³  + 0  x² -   4 x        

                                        -  6  x² +  3 x  -  4

                                        -  6  x²  + 0 x  -  12

                                                        3 x  + 8

Long division:

Therefore, the correct option is "A".

The coordinates of C - (2n, 0) and the coordinates of D - (n, -p). The diagonals of a rhombus are perpendicular and bisect each other.

The properties of a rhombus are:

The given rhombus BCDE has B(n, p) and E(0, 0).

Since the diagonals of a rhombus are perpendicular bisectors,

EO = OC or BO = OD

Where O is the midpoint of EC and BD.

In the given diagram, points B and D are opposite each other. They are reflecting each other over the x-axis.

So, if B has coordinates (n, p) then its reflection over the x-axis is (x, -y) i.e., (n, -p).

Thus, we have B(n, p), D(n, -p), and E(0, 0)

Consider the coordinates of C as (x, y).

The midpoint of BD = ([tex]\frac{n+n}{2}[/tex], [tex]\frac{p-p}{2}[/tex])

⇒ coordinates of O = (n, 0)

So,

The midpoint of EC = ([tex]\frac{0+x}{2}[/tex], [tex]\frac{0+y}{2}[/tex])

⇒ coordinates of O = (x/2, y/2)

⇒ (n, 0) = (x/2, y/2)

∴ x = 2n and y = 0

Then, the coordinates of C are (2n, o)

Therefore, the required coordinates of the given rhombus are C(2n, 0) and D(n, -p).

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Answer:its (-2,1), (-6,-15)

Step-by-step explanation:

Answer: 105

Step-by-step explanation:

105 * .2 = 21 :)

Answer:

Step-by-step explanation:

16 + 12 + 5 + 3 = 36

16 prefer email, 36 total students surveyed

16:36 / 4 = 4:9

4 out of 9 students prefer email

4:9 x 680 = 302.222

302 students can be expected to prefer email.

[tex]s(t) = 17699(1 .066) {}^{t} [/tex]

[tex]s(1) = 17699(1.066) = 18867.13 \\ s(2) = 17699(1.066) {}^{2} = 20112.36 \\ s(5) = 17699(1.066) {}^{5} = 24363.22 \\ s(10) = 17699(1.066) {}^{10} = 33536.73[/tex]

[tex]s(t) = 2 \times initial \: capital \: \\ s(t) = 2 \times 17699[/tex]

[tex]17699(1.066) {}^{t} = 2 (17699) \\ 1.066 {}^{t} = 2 \\ t = log¹°⁰⁶⁶(2) = 10.84511 \: years[/tex]

Answer:

bx/a (3rd choice)

Step-by-step explanation:

The length of an arc, s, in a circle of radius r, and central angle Θ given in radians is

s = rΘ

Circle M has radius a. The central angle in radians of the sector is Θ. The length of the arc is x.

x = aΘ

Solve for Θ:

Θ = x/a      Eq. 1

Circle N has radius b. The central angle in radians of the sector is Θ. The length of the arc is s.

s = bΘ

Solve for Θ:

Θ = s/b      Eq. 2

Since Θ = Θ, then equate the right sides of Equations 1 and 2 above.

x/a = s/b

Multiply both sides by ab.

abx/a = abs/b

bx = as

as = bx

s = bx/a

Answer: bx/a

Step-by-step explanation:

important of festival in nepali languages for class seven

Answer:

b

Step-by-step explanation:

The absolute value equation that satisfies the solution set shown on the number line is given by:

|x| = 1/2

The absolute value function is defined by:

[tex]|x| = x, x \geq 0[/tex]

[tex]|x| = -x, x < 0[/tex]

It measures the distance from a point x to the origin at x = 0. In this problem, the solution set has a distance to the origin of [tex]\frac{1}{2}[/tex], as |-0.5 - 0| = |0.5 - 0| = 0.5, hence the equation is:

|x| = 1/2

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The expression which represents the simplified form of the given expression; x to the 4 power x z to the 5 power over xz to the 6 power as in the task content is; 1/x²z.

According to the task content, it follows that the given expression in the task content is; x⁴z⁵/(xz)⁶.

Hence, the expression can be simplified by means of the laws of indices as follows;

x^(4-6) z^(5-6)

= x-²z-¹

= 1/x²z.

Ultimately, the expression which represents the simplified form of the given expression; x to the 4 power x z to the 5 power over xz to the 6 power as in the task content is; 1/x²z.

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Answer:

35,000 - 8000=27,000 altitude

The vertex form of the quadratic equation, written in standard form, f(x) = 2 · x² - 20 · x + 8 is f(x) + 75 = 2 · (x - 5)².

In this problem we have a quadratic equation in standard form, whose form is defined by f(x) = a · x² + b · x + c, where a, b, c are real coefficients, and we need to transform it into vertex form, defined as:

f(x) - k = C · (x - h)²       (1)

Where:

This latter form can be found by algebraic handling. If we know that f(x) = 2 · x² - 20 · x + 8, then its vertex form is:

f(x) = 2 · x² - 20 · x + 8

f(x) = 2 · (x² - 10 · x + 4)

f(x) + 2 · 25 = 2 · (x² - 10 · x + 25)

f(x) + 75 = 2 · (x - 5)²

The vertex form of the quadratic equation, written in standard form, f(x) = 2 · x² - 20 · x + 8 is f(x) + 75 = 2 · (x - 5)².

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Answer:

Zachary weights 95.7 pounds

Answer: 0

Step-by-step explanation:

Substituting in x=0, we get

[tex]\frac{0(0-2)}{2-2e^{2}(0)}=0[/tex]

Answer:

no solution

Step-by-step explanation:

16x - 32y = 27 → (1)

8x - 16 = 16y ( subtract 16y from both sides )

8x - 16 - 16y = 0 ( add 16 to both sides )

8x - 16y = 16 → (2)

multiply (2) by - 2 and add to (1)

- 16x + 32y = - 32 → (3)

add (1) and (3) term by term

0 + 0 = - 5

0 = - 5 ← not possible

this indicates the system has no solution

(b)

the solution to a system is the point of intersection of the 2 lines

since there is no solution, no point of intersection, then

this indicates the lines are parallel and never intersect

Answer:

a) no solution

b) the two lines never intersect

Step-by-step explanation:

Given system of equations:

[tex]\begin{cases} 16x-32y=27\\8x-16=16y \end{cases}[/tex]

To solve by linear combination (elimination):

Step 1

Write both equations in standard form: Ax + By = C

[tex]\implies 16x-32y=27[/tex]

[tex]\implies 8x-16y=16[/tex]

Step 2

Multiply one (or both) of the equations by a suitable number so that both equations have the same coefficient for one of the variables:

[tex]\implies 16x-32y=27[/tex]

[tex]\implies 2(8x-16y=16) \implies 16x-32y=32[/tex]

Step 3

Subtract one of the equations from the other to eliminate one of the variables:

[tex]\begin{array}{l r l}& 16x-32y & = 32\\- & 16x-32y & = 27\\\cline{1-3}& 0 & =\:\: 5\end{array}[/tex]

Therefore, as 0 ≠ 5, there is no solution to this system of equations.

Part (b)

The solution to a system of equations is the point(s) of intersection.

As there is no solution to the given system of equations, this tells us that the two lines never intersect.

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By the quotient rule for differentiation,

[tex]\displaystyle h'(x) = \dfrac{g(x)f'(x) - f(x)g'(x)}{g(x)^2}[/tex]

According to the plots of [tex]f[/tex] and [tex]g[/tex], we have [tex]f(3)=2[/tex] and [tex]g(3)=3[/tex].

On the interval [2, 4], [tex]f[/tex] is a line through the points (2, 5) and (4, -1), and hence has slope (-1 - 5)/(4 - 2) = -3, so [tex]f'(3)=-3[/tex]. We can similarly find [tex]g'(3)=2[/tex].

Then

[tex]h'(3) = \dfrac{3\cdot(-3)-2\cdot2}{3^2} = \boxed{-\dfrac{13}9}[/tex]

(a) g(- 4) ≈ - 20.566, g(0) = - 8, g(4) = 4, g(6) = 0, g(8) = - 4

(b) g(x) is increasing in the interval [- 4, 2] and decreasing in the interval [4, 8].

(c) There is an up concavity and a decreasing behavior in the interval [2, 6].

(d) The points x = 2 and x = 6 are points of inflection of g(x).

(e) The equation of the line tangent to g(x) at x = 6 is y = - 4 · x + 24.

(f) The range of g(x) is [- 20.566, 4].

(g) The graph of g(x) is shown in the picture attached below.

In this problem we have a piecewise defined function formed by four functions, a circle-like function and three lines, whose integral has to be analyzed in all its characteristics. (a) The integral is described graphically by the area below the curve, where g(2) = 0 and the following properties of the integral are used:

g(- 4) = g(2) - [F(2) - F(- 4)]

g(- 4) = 0 - 0.25π · 4² - 4 · 2

g(- 4) ≈ - 20.566

g(0) = g(2) - [F(2) - F(0)]

g(0) = 0 - 4 · 2

g(0) = - 8

g(4) = g(2) + [F(4) - F(2)]

g(4) = 0 + 0.5 · (2) · (4)

g(4) = 4

g(6) = g(2) + [F(6) - F(2)]

g(6) = 0 + 0.5 · (2) · (4)  - 0.5 · (2) · (4)

g(6) = 0

g(8) = g(2) + [F(8) - F(2)]

g(8) = 0 + 0.5 · (2) · (4) - (2) · (4)

g(8) = - 4

(b) An interval of g(x) is increasing when f(x) > 0 and decreasing when f(x) < 0. Thus, g(x) is increasing in the interval [- 4, 2] and decreasing in the interval [4, 8].

(c) There is an up concavity and a decreasing behavior in the interval [2, 6].

(d) There are points of inflection for values of x such that f'(x) do not exists. The points x = 2 and x = 6 are points of inflection of g(x).

(e) We need to determine the slope and the intercept of the tangent line to determine the equation of the line:

Slope

m = f(6)

m = - 4

Intercept (x = 6, g(x) = 0)

b = g(x) - m · x

b = 0 - (- 4) · 6

b = 24

The equation of the line tangent to g(x) at x = 6 is y = - 4 · x + 24.

(f) The range of g(x) corresponds to the set of values of y that exists in the function. In accordance with the information given in (a), the range of g(x) is [- 20.566, 4].

(g) The graph of g(x) is shown in the picture attached below.

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Answer: yes, as long as you haven't used them. :)

Step-by-step explanation:

Answer:

The no. of student failed is 36.

Step-by-step explanation:

Given, the number of student enrolled= 216

Let us suppose number of student failed = x

Given,

no. of student passed is 5 times no. of student failed.

Then, no. of student passed = 5x

x +5x = 216

6x = 216

x = 216/6

x = 36

Thus, the no. of student failed is 36.

Answer:

56kg

Step-by-step explanation:

If 6m of a rod is 21 kg, then 21/6 will give us the weight of 1m of the rod. 21/6 = 3.5 kg

16*3.5 = 56 kg

Part A - The average value of v(t) over the interval  (0, π/2) is 6/π

Part B -  The displacement of the yo-yo from time t = 0 to time t = π is 0 m

Part C - The total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

The average value of a function f(t) over the interval (a,b) is

[tex]f(t)_{avg} = \frac{1}{b - a} \int\limits^b_a {f(t)} \, dx[/tex]

So, since  velocity at time t is given by v(t) = 3cos(t) for time t ≥ 0. Its average value over the interval  (0, π/2) is given by

[tex]v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {v(t)} \, dt[/tex]

Since v(t) = 3cost, we have

[tex]v(t)_{avg} = \frac{1}{\frac{\pi }{2} - 0} \int\limits^{\frac{\pi }{2} }_0 {3cos(t)} \, dt\\= \frac{3}{\frac{\pi }{2}} \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= \frac{6}{{\pi}} [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= \frac{6}{{\pi}} [{sin(\frac{\pi }{2})} - sin0]\\ = \frac{6}{{\pi}} [1 - 0]\\ = \frac{6}{{\pi}} [1]\\ = \frac{6}{{\pi}}[/tex]

So, the average value of v(t) over the interval  (0, π/2) is 6/π

To find the displacement of the yo-yo, we need to find its position.

So, its position x = ∫v(t)dt

= ∫3cos(t)dt

= 3∫cos(t)dt

= 3sint + C

Given that at t = 0, x = 3. so

x = 3sint + C

3 = 3sin0 + C

3 = 0 + C

C = 3

So, x(t) = 3sint + 3

So, its displacement from time t = 0 to time t = π is

Δx = x(π) - x(0)

= 3sinπ + 3 - (3sin0 + 3)

= 3 × 0 + 3 - 0 - 3

= 0 + 3 - 3

= 0 + 0

= 0 m

So, the displacement of the yo-yo from time t = 0 to time t = π is 0 m

The total distance the yo-yo travels from time t = 0 to time t = π is given by

[tex]x(t) = \int\limits^{\pi}_0 {v(t)} \, dt\\= \int\limits^{\pi }_0 {3cos(t)} \, dt\\= 3 \int\limits^{\pi }_0 {cos(t)} \, dt\\ = 3 \int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt + 3\int\limits^{\pi }_{\frac{\pi }{2}} {cos(t)} \, dt\\= 3 \times 2\int\limits^{\frac{\pi }{2} }_0 {cos(t)} \, dt\\= 6 [{sin(t)}]^{\frac{\pi }{2} }_{0} \\= 6[{sin\frac{\pi }{2} - sin0]\\\\= 6[1 - 0]\\= 6(1)\\= 6[/tex]

So, the total distance the yo-yo travels from time t = 0 to time t = π is 6 m.

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By changing the central angle, ∠CAB, the inscribed angle, ∠CDB has the following data sets:

m∠BAC (β)                                m∠BDC (α)

42°                                               84°

40°                                               80°

45°                                               90°

35°                                               70°

52°                                               104°

A circle can be defined as a closed, two-dimensional curved geometric shape with no edges or corners. Also, a circle refers to the set of all points in a plane that are located at a fixed distance (radius) from a fixed point (central axis).

In Geometry, a circle is considered to be a conic section which is formed by a plane intersecting a double-napped cone that is perpendicular to a fixed point (central axis) because it forms an angle of 90° with the central axis.

The inscribed angle theorem states that the measure of an inscribed angle is one-half the measure of the intercepted arc in a circle. Thus, this is given by this mathematical expression:

m∠BDC = ½ × m∠BAC.

For this exercise, we would change the central angle, ∠CAB, so that the inscribed angle, ∠CDB can have the following data sets:

m∠BAC (β)                                m∠BDC (α)

42°                                               84°

40°                                               80°

45°                                               90°

35°                                               70°

52°                                               104°

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Answer:

plato

Step-by-step explanation:

m∠BAC m∠BDC

50° 25°

70° 35°

90° 45°

125° 62.5°

150° 75°

Answer:

4*1/1000+6*100+8*10+5*1

Step-by-step explanation:

Answer: -4.73, -1.27

Step-by-step explanation:

[tex](x+3)^2 =3\\\\x+3=\pm \sqrt3\\\\\\x=-3 \pm \sqrt3\\\\x \approx -4.73, -1.27[/tex]

Using the z-distribution, the 95% confidence interval for the proportion is given as follows:

(0.4576, 0.5824).

A confidence interval of proportions is given by:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which:

In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.

The other parameters for the interval are given as follows:

[tex]n = 246, \pi = 0.52[/tex].

The lower and upper bound of the interval, respectively, are given by:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 - 1.96\sqrt{\frac{0.52(0.48)}{246}} = 0.4576[/tex]

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 + 1.96\sqrt{\frac{0.52(0.48)}{246}} = 0.5824[/tex]

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The function f(x) = x² then the required function exists g(x) = -x²- 4.

The function exists described as y = f(x).

In mathematics, a function from a set X to a set Y allocates to each element of X exactly one element of Y. The set X exists named the domain of the function and the set Y exists named the codomain of the function. Functions stood originally for the idealization of how a variable quantity relies on another quantity.

For every x there exists a certain value of y.

From the graph g(x) exists reflection of f(x) at y = -4.

So g(x) = -f(x) - 4, the negative sign for reflection.

g(x) = -x² - 4

The required function exists g(x) = -x² - 4.

Therefore, the correct answer is option D. g(x) = -x² - 4.

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The complete question is:

F(x) = x². What is g(x)?

A. g(x) = x² - 4

B. g(x) = x² + 4

C. g(x) = -4x²

D. g(x) = -x² - 4

The expression that will help him to determine the total cost is as follows:

y = 60 + 4x

He signed up to provide snacks for his daughter and her school group's camping trip.

He found a package of granola bars in bulk for $60 that he thought would be enough for everyone for a few days.

He also wanted to buy some boxes of fruit snacks, which were $4 each.

The number to buy and the total cost can be expressed as follows:

let

y = total cost

x = number of boxes of fruit juice.

Therefore, the  expression that will help him to determine the total cost is as follows:

y = 60 + 4x

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