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4 years ago

Use distributive property 9(11x + 12)

1 answer:

6 0

if you use the distributive property, the answer would be 99x+108 because you have to multiple 11x times 9 and 12 times 9

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Snowcat [4.5K]

Hi!

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Area of a triangle = $(base * height)/2$

Area of a square = $(side)^2$

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QUESTION 1: Determine the area of the triangular face.

$\frac{bh}{2}$

$32(20)/2$

$640/2$

$320$

The triangular face is 320 square feet.

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QUESTION 2: Determine the surface area of the entire figure.

$a^2$

$32^2$

$1024$

Now, add the 4 triangular faces to the area of the square face.

$1024+320(4)$

$1024+1280$

$2304$

The total surface area is 1204 square feet.

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For more information, see:

brainly.com/question/4637102

brainly.com/question/22971514

5 0

3 years ago

Need help pls will give you five stars

Temka [501]

Answer:

$\boxed{A. (1,1)}$

Step-by-step explanation:

Well since in,

$f(x) = -3x^2 + 6x - 2$

the 3x^2is negative that means the parabola has a maximum not a minimum meaning we can take out choices,

B and D.

To find the maximum let’s graph the given equation,

Look at the image below.

By looking at the image we can tell that the maximum is at (1,1).

So the answer is choice A (1,1)

3 0

4 years ago

Which equation represents a hyperbola with a center at (0, 0), a vertex at (−48, 0), and a focus at (50, 0)?

Lerok [7]

bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.

$\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1$