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2 years ago

10

What is the measure of YW?

1 answer:

koban [17]2 years ago

8 0

$\huge\boxed{20}$

We know that $\overline{YV}$ and $\overline{VW}$ are equal, and we're looking for their sum. First, we'll need to find the value of $b$ .

Set the expressions equal to each other:

$5b=b+8$

Subtract $b$ from both sides:

$4b=8$

Divide everything by $4$ :

$b=2$

Now that we have the value of $b$ , we can substitute it into the sum of the two expressions to get our final answer.

$5b+(b+8) \\ 5b+b+8 \\ 6b+8$

Substitute:

$6(2)+8$

Simplify:

$12+8 \\ \boxed{20}$

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What is the scale factor in the dilation if the coordinates of A prime are (–7, 6) and the coordinates of C prime are (–4, 3)?

olchik [2.2K]

Option A: $\frac{1}{3}$ is the dilation factor

Explanation:

The Coordinates of the square ABCD are A(-21,18), B(-12,18), C(-12,9), D(-21,9)

Also, the coordinates of the square A'B'C'D' are A'(–7, 6), B'(-4,6), C'(-4,3), D'(-8,3)

Now, we shall determine the scale factor in the dilation if the coordinates A' and C'.

Since, the dilation is the enlargement of the image in the same shape but different size.

To determine the scale factor, let us divide A'C' by AC

Thus, we have,

$\begin{aligned}A^{\prime} C^{\prime} &=\sqrt{(-4+7)^{2}+(3-6)^{2}} \\&=\sqrt{3^{2}+(-3)^{2}} \\&=\sqrt{9+9} \\&=\sqrt{18}\\&=3\sqrt{2} \end{aligned}$

Also,

$\begin{aligned}A C &=\sqrt{(-12+21)^{2}+(9-18)^{2}} \\&=\sqrt{9^{2}+(-9)^{2}} \\&=\sqrt{81+81} \\&=\sqrt{162}\\&=9\sqrt{2} \end{aligned}$

Dividing A'C' by AC, we have,

$\frac{A'C'}{AC} =\frac{3\sqrt{2} }{9\sqrt{2}} =\frac{1}{3}$

Thus, $\frac{1}{3}$ is the dilation factor

3 0

3 years ago

Read 2 more answers

Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v

olga_2 [115]

We are given

$f=z^3 -x^2y$

Firstly, we can find gradient

so, we will find partial derivatives

$f_x=0 -2xy$

$f_x=-2xy$

$f_y=0 -x^2$

$f_y=-x^2$

$f_z=3z^2$

now, we can plug point (-5,5,2)

$f_x=-2*-5*5=50$

$f_y=-(-5)^2=-25$

$f_z=3(2)^2=12$

so, gradient will be

$gradf=(50,-25,12)$

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

so, we will find it's unit vector

$|v|=\sqrt{(-3)^2+(2)^2+(-4)^2}$

$|v|=\sqrt{29}$

now, we can find unit vector

$v'=(\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

now, we can find dot product to find direction of the vector

$dir=(gradf) \cdot (v')$

now, we can plug values

$dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

$dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })$

$dir=-\frac{248\sqrt{29}}{29}$ .............Answer

7 0

3 years ago

Read 2 more answers

If you are doing distance do you add or subtract?

zavuch27 [327]

Answer:

it depends on what the question asks

Step-by-step explanation:

7 0

3 years ago

Can you explain how complex fractions can be used to solve problems involving ratios?

-BARSIC- [3]

Example: (1/5)/(4/13)=(1/5)*(13/4)=13/20

5 0

3 years ago

Please answer this question

Papessa [141]

I don’t get it either

4 0

2 years ago