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igor_vitrenko [27]

3 years ago

10

Find the scale factor of the similar shapes below. For your answer please only input the number. Do not include units with your

answer.

1 answer:

zhenek [66]3 years ago

5 0

Answer:

x is 12

Step-by-step explanation:

7.5÷5=1.5

1.5×8=12

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SVEN [57.7K]

Answer:

2) Equation 1 and Equation 2 have the same number of solutions.

Step-by-step explanation:

The two given equations are

1) 15x + 6 = 41 and 2) 2x + 13 = 28

Solving both equations, we get

Solving (1) : 15x + 6 = 41 ⇒ 15x = 41 - 6 = 35

or, x = 35/15 ⇒ x = 7/3

Solving (2) : 2x + 13 = 28⇒ 2x = 28 - 13 = 15

or, x = 15/2 ⇒ x = 15/2

So, from above solutions we can say that Equation 1 and Equation 2 have the same number of UNIQUE solution.

6 0

3 years ago

Can you graph a line if its slope is undefined? Explain.

JulijaS [17]

You can but it would be going straight up and down

6 0

3 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago

Multiply using the distributive property 12(5y+4).

iragen [17]

Answer:

12(5y+4)

12(9y)

108y

hope it helps and I hope I did it right

Step-by-step explanation:

5 0

3 years ago

8. If BD=BC, BD = 5x -26, BC = 2r + 1, and AC = 43, find AB.

charle [14.2K]

Answer:

24

Step-by-step explanation:

BD=DC

5x-26=2x+1

5x-2x=1+26

3x=27

x=9

BC=2x+1

=2*9+1

BC =19

so AB=AC-BC

AB=43-19

AB=24

8 0

2 years ago