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3 years ago

Please help me!

1 answer:

5 0

4x + 3y = 12
-2x - 5y = 1
multiply bottom by 2
4x + 3y = 12
-4x - 10y = 2
add both equations
-7y = 14
divide by 7
y = -2
plug into first equation
4x + 3(-2) = 12
solve for x
4x = 18
x = 4.5

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the equation -2x^2+3=-9x is rewritten in the form of -2(x-p)^2 +q=0. What is the value of q?the equation -2x^2+3=-9x is rewritte

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Value of q is $\frac{105}{8}$ .

Step-by-step explanation:

We have , the equation $-2x^2+3=-9x$ is rewritten in the form of $-2(x-p)^2 +q=0$ . We need to find What is the value of q . Let's find out:

Let's simplify equation $-2x^2+3=-9x$ into $-2(x-p)^2 +q=0$ :

⇒ $-2x^2+3=-9x$

⇒ $-2x^2+9x+3=0$

⇒ $2x^2-9x=3$

⇒ $2(x^2-\frac{9}{2}x)=3$

⇒ $(x^2-\frac{9}{2}x)=\frac{3}{2}$

⇒ $(x^2-2(1)\frac{9}{4}x)=\frac{3}{2}$

⇒ $(x^2-2(1)\frac{9}{4}x) + (\frac{9}{4})^2=\frac{3}{2} + (\frac{9}{4})^2$

⇒ $(x-\frac{9}{4})^2=\frac{3}{2} + (\frac{81}{16})$

⇒ $(x-\frac{9}{4})^2=\frac{24}{16} + (\frac{81}{16})$

⇒ $(x-\frac{9}{4})^2=\frac{105}{16}$

⇒ $-2(x-\frac{9}{4})^2=-2(\frac{105}{16})$

⇒ $-2(x-\frac{9}{4})^2=-\frac{105}{8}$

⇒ $-2(x-\frac{9}{4})^2+\frac{105}{8} = 0$

Comparing this equation to $-2(x-p)^2 +q=0$ , $q =\frac{105}{8}$ .

Therefore ,Value of q is $\frac{105}{8}$ .

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