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EastWind [94]
2 years ago
9

What is volume of the pyramid??

Mathematics
1 answer:
Bas_tet [7]2 years ago
8 0
For pyramids, the volume is defined by the product of the base area and height divided by 3. Or, V = \frac{Bh}{3} Where B = area of the base and h = the height of the pyramid.

For the figure above:
B = 16 in²
h = 6 in

V =  \frac{16(6)}{3}
V = 32
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Factor 2x2 + 6x – 108.
bekas [8.4K]

Answer:

option A is correct

hope it helps

4 0
2 years ago
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Choose the equation that represents the line that passes through the point (2, 6) and has a slope of −5.
Luden [163]

Answer:

y=-5x+16

Step-by-step explanation:

Plug in the slope and point coordinates into point-slope form (attached in image)

(y-6)=-5(x-2)

1) Distribute -5 to x and -2:

y-6=-5x+10

2) Add 6 to both sides:

y=-5x+16

4 0
3 years ago
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Line p contains point (6, -5) and is perpendicular to line q. The equation for line q is y = 3x + 5. 1) Write an equation for li
wel

Answer:

y = -\frac{1}{3}x -3

Step-by-step explanation:

Perpendicular lines have negative reciprocal slopes. To find the line, convert the slope of q and substitute it with (6, -5) into point slope form.

y = 3x + 5 is q. It has a slope of 3. So p will have a slope of -1/3.

y - y_1 = m(x-x_1)\\y - -5 = -\frac{1}{3}(x - 6)\\y + 5 = -\frac{1}{3}x + 2\\y = -\frac{1}{3}x -3

4 0
2 years ago
I don’t know how to do this
zalisa [80]

Answer:

½÷3=0.17

if you're talking about the app, i don't get too

7 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E%7B2%7D-5x%2B6%7D%7B2x%5E%7B2%7D-7x%2B6%20%7D" id="TexFormula1" title="\frac{x^{
il63 [147K]

Answer:

\frac{x-3}{2x-3}. hole or removable discontinuity at x=2

Step-by-step explanation:

Well generally if you want the simplest form, you factor each the denominator and numerator and then see if you can cancel any of the factors out (because they're in the denominator and numerator)

So let's start by factoring the first equation:

x^2-5x+6

Now let's find what ac is (it's just c since a=1...)

AC= 6

List factors of -6

\pm1, \pm2, \pm3, \pm6.

Now we have to look for two numbers that add up to -5. It's a bit obvious here since there isn't many factors, but it's -2 and -3, and they're both negative since 6 is positive, and -5 is negative...

So using these two factors we get

(x-2)(x-3)

Ok now let's factor the second equation:

2x^2-7x+6

Multiply a and c

AC = 12

List factors of 12:

\pm1, \pm2, \pm3, \pm4, \pm6, \pm12.

Factors that add up to -7 and multiply to 12:

-3\ and\ -4

Rewrite equation:

2x^2-4x-3x+6

Group terms:

(2x^2-4x)+(-3x+6)

Factor out GCF:

2x(x-2)-3(x-2)

Rewrite:

(2x-3)(x-2)

Now let's write out the equation using these factors:

\frac{(x-2)(x-3)}{(2x-3)(x-2)}.

Here we can factor out the x-2 and the simplified form is:

\frac{x-3}{2x-3}

So we can "technically" define f(2) using the most simplified form, but it's removable discontinuity, so it has a hole as x=2. since it makes (x-2) equal to 0 (2-2) = 0.

8 0
1 year ago
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