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butalik [34]
3 years ago
5

Find the volume of the square based pyramid.

Mathematics
2 answers:
labwork [276]3 years ago
5 0
Finding the <span>height of the pyramid:

\left(\dfrac{6}{2} \right )^{\!\!2}+h^{2}=8^{2}\\ \\ \\ 3^{2}+h^{2}=8^{2}\\ \\ h^{2}=8^{2}-3^{2}\\ \\ h^{2}=64-9\\ \\ h^{2}=55\\ \\ h=\sqrt{55}\mathrm{~m}


Finding the base area:

S=6^{2}\\ \\ S=36\mathrm{~m^{2}}


Formula for the volume of the pyramid:

V=\dfrac{S\cdot h}{3}\\ \\ \\ V=\dfrac{36\cdot \sqrt{55}}{3}\\ \\ \\ V=\dfrac{\diagup\!\!\!\! 3\cdot 12\cdot \sqrt{55}}{\diagup\!\!\!\! 3}\\ \\ \\ \boxed{\begin{array}{c} V=12\sqrt{55}\mathrm{~m^{3}} \end{array}}

</span>
frozen [14]3 years ago
4 0
A = 1/3 * s^2 * h where s is a side of the base and h is the height.
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What number can go into 68 and 100
Sati [7]
One number is 4 Another is 2 I hope this helped
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3 years ago
Read 2 more answers
Is 6-6 and 1-6 equal
uranmaximum [27]

Answer:

no

Step-by-step explanation:

6-6=0

1-6=-5

7 0
3 years ago
What is RS? Please help will give Brainlyest.
IceJOKER [234]

Answer:

<h2>         RS = 47</h2>

Step-by-step explanation:

RV=VU and SW=WT    ⇒    VW=\dfrac{RS+UT}{2}

Therefore:

3x+5=\dfrac{2x+15+6x-37}{2}\\\\3x+5=\dfrac{8x-22}{2}

3x + 5 = 4x - 11        {subtract 5 from both sides)

3x = 4x - 16             {subtract 4x from both sides)

-x = -16                   {divide both sides by (-1)}

x=16

So:

RS = 2x + 15 = 2×16 + 15 = 32 + 15 = 47

3 0
2 years ago
Solve the system:<br> x + 5y = -15<br> -2x -10y = 30*
Ksju [112]

Answer:

x= -5/3 y-5

Step-by-step explanation:

6 0
3 years ago
Select interior, exterior, or on the circle (x - 5) 2 + (y + 3) 2 = 25 for the following point. (2, 3)
lozanna [386]
Standard equation of a circle: <em>(x-h)² + (y-k)² = r²</em> where <em>(h, k)</em> is the center and <em>r </em>is the radius. In the case of our equation here, <em>(x-5)² + (y+3)² = 25</em>, we can conclude that our circle has a center at (5, -3) and a radius of 5 units.

We can use the distance formula with the center (5, -3) and our point (2, 3) to see how far away they are...if the distance between them is less than the radius of the circle, it is on the interior. If it's equal, it's on the circle. If it's greater, it's on the exterior.

Distance = \sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

Distance = \sqrt{(-3-3)^2+(5-2)^2}

Distance = \sqrt{(-6)^2+3^2}

Distance = \sqrt{36+9}

Distance = \sqrt{45}\approx6.7082

6.7082>5,\ so\ (2,3)\ is\ on\ the\ \boxed{exterior}
8 0
3 years ago
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