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marishachu [46]

3 years ago

10

Allie noticed that to get her test grade, she could take Valerie's grade, multiply it by 2 and subtract 15. If Valerie's test gr

ade was x, how would you write Allie's test grade? A) 2x B) 2x + 15 C) 2x - 15 D) 15x - 2

1 answer:

kolbaska11 [484]3 years ago

4 0

Answer is c) 2x - 15 because x is Valerie's grade

Allie's grade is <span>taking Valerie's grade, multiply it by 2 and subtract 15</span>

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Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

likoan [24]

Answer:

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Step-by-step explanation:

Consider, the given Quadratic equation, $x^2+4=6x$

This can be written as , $x^2-6x+4=0$

We have to solve using quadratic formula,

For a given quadratic equation $ax^2+bx+c=0$ we can find roots using,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ...........(1)

Where, $\sqrt{b^2-4ac}$ is the discriminant.

Here, a = 1 , b = -6 , c = 4

Substitute in (1) , we get,

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1 \cdot (4)}}{2 \cdot 1}$

$\Rightarrow x=\frac{6\pm\sqrt{20}}{2}$

$\Rightarrow x=\frac{6\pm 2\sqrt{5}}{2}$

$\Rightarrow x={3\pm \sqrt{5}}$

$\Rightarrow x_1={3+\sqrt{5}}$ and $\Rightarrow x_2={3-\sqrt{5}}$

We know $\sqrt{5}=2.23607$ (approx)

Substitute, we get,

$\Rightarrow x_1={3+2.23607}$ (approx) and $\Rightarrow x_2={3-2.23607}$ (approx)

$\Rightarrow x_1={5.23607}$ (approx) and $\Rightarrow x_2=0.76393}$ (approx)

Thus, the two root of the given quadratic equation $x^2+4=6x$ is 5.24 and 0.76 .

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We can also solve it.

3x-9 = 6x6/4

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Step-by-step explanation:

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