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3 years ago

9

Solve with completing the square method:6x²-7x+2=0 and ax²+bx+c=0

1 answer:

Norma-Jean [14]3 years ago

7 0

$6x ^2-7x+2=0\\ \\ a=6 , \ b = -7 , \ c=2 \\ \\\Delta = b^{2}-4ac = (-7)^{2}-4*6*2=49-48=1 \\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{7-\sqrt{1}}{2*6}=\frac{7-1}{12} =\frac{6}{12}= \frac{1}{ 2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{7+\sqrt{1}}{2*6}=\frac{7+1}{12} =\frac{8}{12}= \frac{2}{ 3}$

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Please Show your work.

lukranit [14]

Answer:

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Step-by-step explanation:

Equation: 10 - 7.5 = 2.5 and when you subtract 2.5 until the Number of Drawers (x) is 0, then you get 0 for Total Volume (y), so y =2.5x+0 or y = 2.5x

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dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

<u>Area of the big rectangle</u>: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

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2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

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

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It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

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$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

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$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

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3 years ago