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4 years ago

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The vertex of a parabola is at point 2,5.Which equations could be for this parabola below??

1 answer:

Natali5045456 [20]4 years ago

7 0

The vertex form of the equation of a vertical parabola is

y = (x-h)^2 + k, where (h,k) is the vertex.

Here, because the vertex is at (2,5), that equation must be:

y = (x-2)^2 + 5.

Answer A is closest, but ought to be A) y = (x-2)^2 + 5 (the 5 is positive).

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4 years ago

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base

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Answer:

The base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = $x\times\frac{1}{2} =\frac{x}{2} \ m.$

The height of triangle = $x\times \frac{3}{4} =\frac{3x}{4}\ m.$

The area of triangle = $16\ m^2.$

Now, we put the formula of area to solve:

$Area=\frac{1}{2} \times base\times height$

$16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}$

$16=\frac{3x^2}{16}$

<em>Multiplying both sides by 16 we get:</em>

<em /> $256=3x^2$ <em />

<em>Dividing both sides by 3 we get:</em>

<em /> $\frac{256}{3} =x^2$ <em />

<em>Using square root on both sides we get:</em>

$\frac{16}{\sqrt{3}}=x$

$x=\frac{16}{\sqrt{3}}$

Now, by substituting the value of $x$ to get the base and height:

$Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.$

<em>So, the base of triangle = </em> $\frac{8}{\sqrt{3}} \ m.$ <em />

$Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.$

<em>Thus, the height of triangle = </em> $\frac{12}{\sqrt{3}} \ m.$ <em />

Therefore, the base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

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