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2 years ago

What set of transformations are applied to parallelogram ABCD to create A'B'C'D'?

Mathematics

2 answers:

iogann1982 [59]2 years ago

4 0

Answer:

Reflected over the x‒axis and reflected over the y-axis

Step-by-step explanation: pls mark me as brainly

denpristay [2]2 years ago

3 0

Answer:

Reflected over the x-axis and rotated 90° counterclockwise

Step-by-step explanation:

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2 years ago

Read 2 more answers

A student solves the following equation and

Phoenix [80]

Answer:

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

$\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}$

$\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}$

$\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)$

$\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)$

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)$

$\frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)$

$-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a$

$\frac{1}{a-2}\left(a+2\right)\left(a-2\right):\quad a+2$

so equation becomes

$3\left(a-2\right)-6a=a+2$

$-3a-6=a+2$

$-3a-6+6=a+2+6$

$-4a=8$

$\mathrm{Divide\:both\:sides\:by\:}-4$

$\frac{-4a}{-4}=\frac{8}{-4}$

$a=-2$

$\mathrm{Verify\:Solutions}$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}$

$\mathrm{Solve\:}\:a+2=0:\quad a=-2$

$\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2$

$\mathrm{Solve\:}\:a-2=0:\quad a=2$

So the following points are undefined

$a=-2,\:a=2$

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

4 0

3 years ago

Write the quadratic equation whose roots are 3 and -5, and whose leading coefficient is 2.

Rainbow [258]

$\begin{cases} x=3\implies &x-3=0\\ x=-5\implies &x+5=0 \end{cases}\qquad \implies\qquad a(x-3)(x+5)=\stackrel{0}{y} \\\\\\ a(\stackrel{F~O~I~L}{x^2+2x-15})=y\implies \stackrel{\textit{leading coefficient of 2}}{2(x^2+2x-15)}=y\implies 2x^2+4x-30=y$

Check the picture below.

5 0

2 years ago