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Veronika [31]
3 years ago
9

Can someone help me with these math problems please? I need it done by tonight if someone can please help me by then...I'll make

the one with the right answers brainliest. Don't answer just for the points.. So if anyone can please help me it'll be helpful please and thanks must have right answers and done by tonight..

Mathematics
1 answer:
kotegsom [21]3 years ago
4 0

Answer:

For question 3, you would just add 2 to the x values and subtract 2 from the y values, so it would be:

J' (-2, 5)

K' (2, 6)

L' (1, 2)

M' (-3, 1)

For question 4 you would subtract 7 from the x values and 6 from the y values, and that would be:

W' (-6, 1)

X' (-1, -1)

Y' (-3, -6)

Z' (-8, -4)

For question 9 you would end up with:

X' (6, -5)

Y' (7, 1)

Z' (4, 0)

For question 10 you would end up with:

Q' (-1, 2)

R' (1, 7)

S' (-2, 6)

T' (-4, 1)

For question 11 you would end up with:

L' (4, 1)

M' (8, 5)

N' (6, 7)

P' (2, 3)

For question 12 you would end up with:

G' (6, -7)

H' (6, -4)

I' (1, -7)

Hope this is what you were looking for!

 

Step-by-step explanation:


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Solve. -7x+1-10x^2=0
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-7x+1-10x^2=0

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quadratic\:equation:- ax^2+bx+c=0

solutions:-\\\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

For \\A=-10\\B=-7\\C=1

x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{\left(-7\right)^2-4\left(-10\right)\cdot \:1}}{2\left(-10\right)}

\sqrt{\left(-7\right)^2-4\left(-10\right)\cdot \:1}=\sqrt{89}

x_{1,\:2}=\frac{-\left(-7\right)\pm \sqrt{89}}{2\left(-10\right)}

x_1=\frac{-\left(-7\right)+\sqrt{89}}{2\left(-10\right)},\:x_2=\frac{-\left(-7\right)-\sqrt{89}}{2\left(-10\right)}

\frac{-\left(-7\right)+\sqrt{89}}{2\left(-10\right)}=-\frac{7+\sqrt{89}}{20}

\frac{-\left(-7\right)-\sqrt{89}}{2\left(-10\right)}=\frac{\sqrt{89}-7}{20}

x=\frac{\sqrt{89}-7}{20}

<u>OAmalOHopeO</u>

<u />

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