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3 years ago

Which transformation(s) of circle C prove that circle A is similar to circle C ?

Mathematics

1 answer:

Bas_tet [7]3 years ago

6 0

It similar Because there both in the 2s

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How to solve inequalities with word problems

s2008m [1.1K]

1. Read through the entire problem. 2. Highlight the important information and key words that you need to solve the problem. 3. Identify your variables. 4. Write the equation or inequality. 5. Solve. 6. Write your answer in a complete sentence. 7. Check or justify your answer. hope this helps :D

7 0

3 years ago

4(2x - 7) - 2x = -10

jarptica [38.1K]

So the right answer is 3.

look at the attached picture

Hope it will help you..

Good luck on your assignment

~pragya

5 0

3 years ago

Read 2 more answers

2. G(x) = 2x² + 3x Is the function even, odd, or neither?

Bingel [31]

The function appears to be neither odd nor even

8 0

3 years ago

X2 - 6x + 12 and y = 2x - 4, algebraically are

olga_2 [115]

The solution is (4, 4)

Solution:

Given system of equations are:

$y = x^2 - 6x + 12 ------ eqn\ 1\\\\y = 2x - 4 ---------- eqn\ 2$

Substitute eqn 2 in eqn 1

$x^2 - 6x + 12 = 2x - 4$

Make the right side of equation 0

$x^2 - 6x + 12 - 2x + 4 = 0\\\\x^2 -8x + 16 = 0$

Solve by quadratic equation

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=-8,\:c=16:\\\\x=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:16}}{2\cdot \:1}\\\\x=\frac{-\left(-8\right)\pm \sqrt{0}}{2\cdot \:1}\\\\x = \frac{8}{2}\\\\x = 4$

Substitute x = 4 in eqn 2

y = 2(4) - 4

y = 8 - 4

y = 4

Thus solution is (4, 4)

5 0

3 years ago

Pls answer this One way to solve an equation or inequality that has both decimals and fractions is to change the

KatRina [158]

Answer:

Turn everything into fractions. You can also combine like terms.

Step-by-step explanation:

5 0

2 years ago