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3 years ago

14

How do you do dilation ???

1 answer:

charle [14.2K]3 years ago

7 0

Starting with ΔABC, draw the dilation image of the triangle with a center at the origin and a scale factor of two. Notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). Dilation with scale factor 2, multiply by 2.

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#9 Please. 20 points. I keep getting x is 25 or x 12.25 depending on how I do it

Juliette [100K]

Answer:

3x+10=(2x-17)*2

(I multiplied it by 2, since ”in circle theorems”, she angle sub tended at the center, is 2* the angle sub tended at the circumference)

3x+10=4x-34

-x = -44

x = 44

8 0

4 years ago

What is the value of the expression below when x=5?

vredina [299]

5 =x that is the expression

7 0

3 years ago

Factor over complex numbers 2x^4+36x^2+162

Darina [25.2K]

<u>Answer:</u>

Factor over complex numbers of $2 x^{4}+36 x^{2}+162$ is $2(x+3 i)^{2}(x-3 i)^{2}$

<u>Solution: </u>

From question given that

$2 x^{4}+36 x^{2}+162$ → (1)

On substituting $x^{2}=y$ in equation (1),

$=2 y^{2}+36 y+162$

Taking 2 as a common in above expression,

$=2\left(y^{2}+18 y+81\right)$

Rewrite the above expression,

$=2\left[y^{2}+2(y)(9)+9^{2}\right]$

$=2(y+9)^{2}$ $\left[\text { Using }(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

$=2\left(x^{2}+9\right)^{2}$ $\left[\text { since } y=x^{2}\right]$

$\left[\text { Using } a^{2}+b^{2}=(a+i b)(a-i b), \text { where } i=\sqrt{-1}\right]$

$=2[(x+3 i)(x-3 i)]^{2}$

$=2(x+3 i)^{2}(x-3 i)^{2}$

Hence Factor over complex numbers of $2 x^{4}+36 x^{2}+162$ is $2(x+3 i)^{2}(x-3 i)^{2}$

7 0

4 years ago

Write a formula for the maximum number of segments determined by n points

jeyben [28]

Let n = the number of points
(x-1) + ... (x-x)

The last term will always be 0, when you reach that, stop.

ex. 1pt: 1-1=0
2pt: (2-1) + (2-2) = 1
3pt: (3-1) + (3-2) + (3-3) = 3

7 0

3 years ago

Read 2 more answers

Q # 14 please . .Write the slope-intercept of the equation for the line

Valentin [98]

You can see that this line passe through the points (-4,-2) and (4,3).

The slope of the line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is

$\dfrac{y_2-y_1}{x_2-x_1}.$

Thus, the slope is

$\dfrac{3-(-2)}{4-(-4)}=\dfrac{5}{8}.$

Then the equation of the line is

$y-3=\dfrac{5}{8}(x-4),\\\\y=\dfrac{5}{8}x+\dfrac{1}{2}.$

Answer: correct choice is C.

7 0

4 years ago