What is the distance from (-3,8) and (13,-6)?

Answer:

t2=8/5

Step-by-step explanation:

using this formula

t1/s1 =t2/s2

4/5=t2/2

cross multiply

5t2=8

t2=8/5

The correct answer for the value of t₂ is [tex]1.6[/tex].

Given:

Time t₁ = 4,

Distance s₂ =2

Distance s₁ = 5.

To find value of t₂ , use the concept of proportion:

[tex]\dfrac{t_1}{s_1} = \dfrac{t_2}{s_2}[/tex]

Put value of [tex]t_1 ,s_1 ,s_2[/tex]:

[tex]\dfrac{t_2}{2} =\dfrac{4}{5}\\\\t_2 =\dfrac{8}{5}\\\\ t_2 = 1.6[/tex]

The correct value of [tex]t_2[/tex] is [tex]1.6[/tex].

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Step-by-step explanation:

21) f(x)=1/x-6. g(x)=7/x+6

f(g(x))=f(7/x+6)=1÷7/x+6 - 6=x+6/7 - 6

g(f(x))=g(1/x-6)=7÷1/x-6 - 6 =7(x-6) - 6

simplify forward

25)f(x)=|x| g(x)=5x+1

f(g(x))=f(5x+1)=|5x+1|=5x+1=g(x)

g(f(x))=g(|x|)=5|x|+1=5x+1=g(x)

Answer:  d: 1:52pm

Step-by-step explanation:  Since Beatrice finished 22 minutes earlier, we subtract 22 minutes from 2:14. 2:14 - 14 is 2:00. 22-14 is 8. 2:00 - 8 is 1:52.

Answer: 10,204 feet

Step-by-step explanation: i would assume you would add the two together, seeing as if the mountain is 10,093 above sea level and the valley is 111 below, the difference in elevation is also the distance between each other.

The lowest common multiple of the expressions 3xyz^2 and 9x^2y + 9x^2 is 9x^2z^2(y + 1)

The expressions are given as:

3xyz^2 and 9x^2y + 9x^2

Factorize the expressions

3xyz^2 = 3 * x * y * z * z

9x^2y + 9x^2 = 3 * 3 * x * x * (y + 1)

Multiply the common factors, without repetition

LCM = 3 * 3 * x * x * (y + 1) * z* z

Evaluate the product

LCM = 9x^2z^2(y + 1)

Hence, the lowest common multiple of the expressions 3xyz^2 and 9x^2y + 9x^2 is 9x^2z^2(y + 1)

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(i) The equation of the axis of symmetry is x = - 5.

(ii) The coordinates of the vertex of the parabola are (h, k) = (4, - 18). The x-value of the vertex is 4.

(iii) According to the vertex form of the quadratic equation, the parabola opens down due to negative lead coefficient and has a vertex at (2, 4), which is a maximum.  

In this question we must find and infer characteristics from three cases of quadratic equations. (i) In this case we must find a formula of a axis of symmetry based on information about the vertex of the parabola. Such axis passes through the vertex. Hence, the equation of the axis of symmetry is x = - 5.

(ii) We need to transform the quadratic equation into its vertex form to determine the coordinates of the vertex by algebraic handling:

y = x² - 8 · x - 2

y + 18 = x² - 8 · x + 16

y + 18 = (x - 4)²

In a nutshell, the coordinates of the vertex of the parabola are (h, k) = (4, - 18). The x-value of the vertex is 4.

(iii) Now here we must apply a procedure similar to what was in used in part (ii):

y = - 2 · (x² - 4 · x + 2)

y - 4 = - 2 · (x² - 4 · x + 2) - 4

y - 4 = - 2 · (x² - 4 · x + 4)

y - 4 = - 2 · (x - 2)²

According to the vertex form of the quadratic equation, the parabola opens down due to negative lead coefficient and has a vertex at (2, 4), which is a maximum.  

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Answer:

Step-by-step explanation:

The correct value of w is 20 as the width of a rectangle must be positive. A quadratic function always has 2 zeroes and in a case like this the negative one is ignored.

Answer

{-10, 9}

Step-by-step explanation:

The above are Simultaneous equations

These are two or more equations that share same variables.

8X + 10Y = 10--------(1)

5X + 4Y = -14-------(2)

40X + 50Y = 50 ---------(3)

40X + 32Y = - 112--------(4)

18Y = 162

[tex] \frac{18y}{18} = \frac{162}{18} \\ y = 9[/tex]

40X + 50Y = 50

40X + 50(9) = 50

40X + 450 = 50

40X = 50 - 450

40X = - 400

{-10, 9}

Rewriting the left-hand side as follows,

[tex](2\cos\theta-6\sin \theta)^2 +(6\cos \theta+2\sin \theta)^2\\\\=4\cos^2 \theta-24\cos \theta \sin \theta+36 \sin^2 \theta+36 \cos^2 \theta+24 \cos \theta \sin \theta+4 \sin^2 \theta\\\\=40\cos^2 \theta+40 \sin^2 \theta\\\\=40(\cos^2 \theta+\sin^2 \theta)\\\\=40[/tex]

Answer:

  x = 23; not extraneous

Step-by-step explanation:

A solution is extraneous if it does not satisfy the original equation. Extraneous solutions can sometimes be introduced in the process of solving radical and rational function equations.

Squaring both sides of the given equation, we get ...

  √(3x +12) = 9

  3x +12 = 81 . . . . . . square both sides

  x +4 = 27 . . . . . . . divide by 3

  x = 23 . . . . . . . . . . subtract 4

There is only one solution, and it satisfies the equation:

  √(3×23 +12) = √81 = 9

The solution x = 23 is not extraneous.

Answer: x = 23; not extraneous

Step-by-step explanation:

Answer: c) {1,245; 1,224; 1,203; 1,182; …}

Step-by-step explanation:

Concept:

For this question, we just go by eliminating each answer until we get the correct one

Given information

Common difference = -21 (decreasing sequence)

Answer Choice: a) {873, 894, 915, 936, …}

894 - 873 = 21

915 - 895 = 21

936 - 915 = 21

Since the common difference is 21, not -21

[tex]\large\boxed{FALSE}[/tex]

Answer Choice: b) {32, 20, 8, -4, …}

20 - 32 = -12

8 - 20 = -12

-4 - 8 = -12

Since the common difference is -8, not -21

[tex]\large\boxed{FALSE}[/tex]

Answer Choice: c) {1,245; 1,224; 1,203; 1,182; …}

1224 - 1245 = -21

1203 - 1224 = -21

1182 - 1203 = -21

Since the common difference is -21

[tex]\Huge\boxed{TRUE}[/tex]

Answer Choice: d) {1,563; 1,587; 1,611; 1,635; …}

1587 - 1563 = 24

1611 - 1587 = 24

1635 - 1611 = 24

Since the common difference is 24, not -21

[tex]\large\boxed{FALSE}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer: $2250

Step-by-step explanation:

Given information

Dimension of the room = 20 feet × 15 feet

Cost of tiles = $7.50 / ft²

Derived formula from the given information

Total cost = Floor area × Cost of tiles

Substitute values into the formula

Total cost = Floor area × Cost of tiles

Total cost = (20 × 15) × (7.5)

Simplify by multiplication

Total cost = 300 × (7.5)

[tex]\Large\boxed{Total~cost~=~2250~Dollars}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

Consider the table for y= |x+1| :

x   |   y

---------

0      1

1       2

2      3

-1      2

-2     3

This would give us the parent function of y=|x| but translated up one unit. It should look like a v starting at (0, 1)

Given the height of the tree and the angle of elevation from point B, the distance between the tree is from point B is approximately 70ft.

Given the data in the question;

Since the scenario form a right angle triangle, we use trig ratio.

tanθ = Opposite / Adjacent

tan( 26° ) = 34ft / x

We solve for x
x = 34ft / tan( 26° )

x = 34ft / 0.4877

x = 70ft

Given the height of the tree and the angle of elevation from point B, the distance between the tree is from point B is approximately 70ft.

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Neil had 18 dimes, 6 nickels and 10 quarters saved up in his jar making a total of 460 pennies

An equation is an expression that shows the relationship between two or more numbers and variables.

In his piggy bank, Neil has three times as many dimes as nickels and he has four more quarters than nickels. The coins total are 4.60. How many of each coin does he have.

Let x represent dimes (0.1), y represent nickels (0.5) and z represent quarter (0.25)

Hence:

0.1x + 0.05y + 0.25z = 4.6   (1)

Also:

x = 3y   (2)

And:

z = y + 4  (3)

From the equations:

x = 18, y = 6, z = 10

Neil had 18 dimes, 6 nickels and 10 quarters saved up in his jar making a total of 460 pennies

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Answer:

See below for proof.

Step-by-step explanation:

Given:

[tex]y=\left(x+\sqrt{1+x^2}\right)^m[/tex]

First derivative

[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}[/tex]

[tex]\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}[/tex]

Second derivative

[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]

[tex]\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}[/tex]

[tex]\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}[/tex]

[tex]\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m[/tex]

[tex]\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m[/tex]

[tex]\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}[/tex]

              [tex]= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)[/tex]

Proof

  [tex](x^2+1)y_2+xy_1-m^2y[/tex]

[tex]= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m[/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m[/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right][/tex]

[tex]= \left(x+\sqrt{1+x^2}\right)^m\left[0][/tex]

[tex]= 0[/tex]

The solutions for the given equations are:

Writing a number or an equation as a product of its factors is said to be the factorization.

A linear equation has only one factor, a quadratic equation has 2 factors and a cubic equation has 3 factors.

1. Solving x² - 49 = 0; (quadratic equation)

⇒ x² - 7² = 0

This is in the form of a² - b². So, a² - b² = (a + b)(a - b)

⇒ (x + 7)(x - 7) =0

By the zero-product rule,

x = -7 and 7.

2. Solving 3x³ - 12x = 0

⇒ 3x(x² - 4) = 0

⇒ 3x(x² - 2²) = 0

⇒ 3x(x + 2)(x - 2) = 0

So, by the zero product rule, x = -2, 0, 2

3. Solving 12x² + 14x + 12 = 18; (quadratic equation)

⇒ 12x² + 14x + 12 - 18 = 0

⇒ 12x² + 14x - 6 = 0

⇒ 2(6x² + 7x - 3) = 0

⇒ 6x² + 9x - 2x - 3 = 0

⇒ 3x(2x + 3) - (2x + 3) = 0

⇒ (3x - 1)(2x + 3) = 0

∴ x = 1/3, -3/2

4. Solving -x³ + 22x² - 121x = 0

⇒ -x³ + 22x² - 121x = 0

⇒ -x(x² - 22x + 121) = 0

⇒ -x(x² - 11x - 11x + 121) = 0

⇒ -x(x(x - 11) - 11(x - 11)) = 0

⇒ -x(x - 11)² = 0

∴ x = 0, 11, 11

5. Solving x² - 4x = 5; (quadratic equation)

⇒ x² - 4x - 5 = 0

⇒ x² -5x + x - 5 = 0

⇒ x(x - 5) + (x - 5) = 0

⇒ (x + 1)(x - 5) =0

∴ x = -1, 5

Hence all the given equations are solved.

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Answer:

ratio = 4

Step-by-step explanation:

According to the given table:

• the number of days with no fire incidents

  = 16

• the number of days with more than 5 fire incidents

  = 2 + 2

  = 4

Conclusion :

the ratio of the number of days with no fire incidents

to the number of days with more than 5 fire incidents is :

16 to 4 (16 : 4)

Then

The ratio = 4

Answer:

yes

Step-by-step explanation:

25x² - 40xy + 16y² can be factored as

(5x - 4y)² ← a perfect square

The psi that the TPMS would trigger a warning for this car is = 23.36 psi

The target tire pressure of the car is = 32 psi (pounds per square inch.)

The Tire pressure monitoring systems (TPMS) warns the car below 27% of 32psi

That is , 27/100 × 32

= 864/100

= 8.64psi

Therefore, 32 - 8.64 = 23.36. When the car is below 23.36psi, TPMS would trigger a warning for this car.

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Complete question:

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 27% below the target pressure. Suppose the target tire pressure of a certain car is 32 psi (pounds per square inch.)

At what psi will the TPMS trigger a warning for this car? (Round your answer to 2 decimal place.) When the tire pressure is above or below?

The Pythagorean theorem states that:

[tex]a^2+b^2=c^2[/tex]

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

Let a represent the length of one leg.

Because the hypotenuse is 14 cm longer than a leg, we can say that the hypotenuse's length is 14 + a.

Plug these into the Pythagorean theorem:

[tex]a^2+b^2=c^2\\a^2+a^2=(14+a)^2\\2a^2=14^2+2(14)a+a^2\\2a^2=196+28a+a^2\\a^2=196+28a\\a^2-196-28a=0\\a^2-28a-196=0[/tex]

Factor using the quadratic formula:

[tex]a=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]a=\dfrac{-(-28)\pm \sqrt{(-28)^2-4(1)(-196)}}{2(1)}\\\\a=14\pm14\sqrt{2}\\\\a=14+14\sqrt{2}[/tex]

We know that it's plus because subtracting results in a negative value, and length cannot be negative.

This is the length of each side.

Because the hypotenuse is 14 cm longer, we can say that the hypotenuse is [tex]28+14\sqrt{2}[/tex].

Leg length = [tex]14+14\sqrt{2}[/tex]

Hypotenuse length = [tex]28+14\sqrt{2}[/tex]

If the formula, =MID("ABCDEFGHI",3,4) is used, the result yielded would be CDEF.

When using the =MID function on a spreadsheet, the number after the text in the formula would show the position of the text from the left that the function would begin to count from.

The text in the third position from the left as shown in ABCDEFGHI is C so we need to start counting from letter C.

The second number in the function would then show the number of texts that needs to be counted and selected from the row of letters. That number is 4.

So from the letter C, you'll count 4 letters including the letter C itself.

The result you get would therefore be C, D, E, F which are the four letters from C.

In conclusion, the formula =MID("ABCDEFGHI",3,4) would yield the result, "C, D, E, F,."

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Triangle 1, triangle 3 and triangle 4 are congruent triangles bases on side-side-side and side-angle-side congruency.

Triangle is a polygon that has three sides and three angles. Types of triangles are isosceles, equilateral and scalene triangle.

Two triangles are said to be congruent if they have the same shape and their corresponding sides are congruent to each other. Also, their corresponding angles are congruent.

Triangle 1, triangle 3 and triangle 4 are congruent triangles bases on side-side-side and side-angle-side congruency.

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Usando la distribución hipergeométrica,  la probabilidad de que ambas estén numeradas con el valor 1, es: 0.0222 = 2.22%.

La fórmula es:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Los parámetros son:

Los valores de los parámetros son:

N = 10, k = 2, n = 2.

La probabilidad de que ambas estén numeradas con el valor 1, es P(X = 2), entonces:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,10,2,2) = \frac{C_{2,2}C_{8,0}}{C_{10,2}} = 0.0222[/tex]

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Answer:

Step-by-step explanation:

Angles is known to be opposite each other when two lines cross called vertical angles  due to the fact that they are opposite each other at a vertex.

Vertical Angles are known to be often called Vertically Opposite Angles and this is described as the scenario when two lines intersect one another, then the opposite angles, is made as a result of the intersection which is known to be called vertical angles or what we say as vertically opposite angles.

Note that A pair of vertically opposite angles are said to be often always equal to one another.

Hence, based on the scenario above, Angles is known to be opposite each other when two lines cross called vertical angles  due to the fact that they are opposite each other at a vertex.

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[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Let's solve ~

Equation of directrix is : y = 1, so we can say that it's a parabola of form : -

[tex]\qquad \sf  \dashrightarrow \: (x - h) {}^{2} = 4a(y - k)[/tex]

and since the focus is above the directrix, it's a parabola with upward opening.

[tex]\qquad \sf  \dashrightarrow \: (x - ( - 4)) {}^{2} = 4(2)(y - 5)[/tex]

[tex]\qquad \sf  \dashrightarrow \: (x + 4) {}^{2} = 8(y - 5)[/tex]

[tex]\qquad \sf  \dashrightarrow \: {x}^{2} + 8x + 16 = 8y - 40[/tex]

[tex]\qquad \sf  \dashrightarrow \: 8y = {x}^{2} + 8x + 56[/tex]

[tex]\qquad \sf  \dashrightarrow \: y = \cfrac{1}{8} {x}^{2} + x + 7[/tex]

Directrix

Focus

Focus lies in Q3 and above y=1

Then

Perpendicular distance

Find a for the equation

Now the equation is

[tex]\\ \rm\dashrightarrow 4a(y-k)=(x-h)^2[/tex]

[tex]\\ \rm\dashrightarrow 4(2)(y-5)=(x+4)^2[/tex]

[tex]\\ \rm\dashrightarrow 8(y-5)=x^2+8x+16[/tex]

[tex]\\ \rm\dashrightarrow 8y-40=x^2+8x+16[/tex]

[tex]\\ \rm\dashrightarrow 8y=x^2+8x+16+40[/tex]

[tex]\\ \rm\dashrightarrow 8y=x^2+8x+56[/tex]

[tex]\\ \rm\dashrightarrow y=\dfrac{x^2}{8}+x+7[/tex]

An angle is produced at the point where two or more lines meet. Thus the value of LP required in the question is approximately 14.

Two lines are said to be perpendicular when a measure of the angle between them is a right angle. While parallel lines are lines that do not meet even when extended to infinity.

From the question, let the length of LP be represented by x.

Thus, from the given question, it can be deduced that;

LM ≅ PN = 20

KP = KN - PN

    = 30 - 20

KP = 10

LP = x

Also,

<MLP is a right angle, so that;

< KLP = < KLM - <PLM

         = 126 - 90

<KLP = [tex]36^{o}[/tex]

So that applying the Pythagoras theorem to triangle KLP, we have;

Tan θ = [tex]\frac{opposite}{adjacent}[/tex]

Tan 36 = [tex]\frac{10}{x}[/tex]

x = [tex]\frac{10}{Tan 36}[/tex]

  = [tex]\frac{10}{0.7265}[/tex]

x = 13.765

Therefore the side LP ≅ 14.

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The vertices of the feasible region are (0, -2), (-3, 1) and (3, 7)

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

(-3, 1), (3, 7) and (0, -2)

So, we have:

(0, -2)

(-3, 1)

(3, 7)

Hence, the vertices of the feasible region are (0, -2), (-3, 1) and (3, 7)

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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