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ivolga24 [154]
3 years ago
15

Which describes the cross section of a square prism that passes through vertices A, B, and C?

Mathematics
2 answers:
expeople1 [14]3 years ago
6 0

Answer:

In triangle ABC one side equal 8\sqrt{2} and two sides equal 4\sqrt{13}

Step-by-step explanation:

We are given a prism whose base is square with sides 8 in and height 12 in.

If we take cross section through vertices A, B and C

We will get a cross section as triangle.

In triangle ABC, sides are AB, BC and AC

AB is diagonal of top square whose side 8 in.

AB=\sqrt{8^2+8^2}=8\sqrt{2}

AC is face diagonal of front face.

AC=\sqrt{8^2+12^2}=4\sqrt{13}

BC is face diagonal of right face.

BC=\sqrt{8^2+12^2}=4\sqrt{13}

AC=BC≠AB

Hence, In triangle ABC one side equal 8\sqrt{2} and two side equal 4\sqrt{13}

CaHeK987 [17]3 years ago
3 0

Answer:

A) A recangle that is not a square

Step-by-step explanation:

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If you started the problem by converting 1 pound to 16 ounces, would you get the same results?
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Answer:

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Step-by-step explanation:

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Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the
Rama09 [41]

<em><u>Question:</u></em>

Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?

1s   2s   3s   4s

<em><u>Answer:</u></em>

It takes 2 seconds for object to hit the ground

<em><u>Solution:</u></em>

<em><u>The given equation is:</u></em>

h(t) = -16t^2 + v_0t+h_0

Initial velocity = 27 feet/sec

h_0 = 10\ feet

Therefore,

h(t) = -16t^2 +27t+10

At the point the object hits the ground, h(t) = 0

-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0

Solve by quadratic formula,

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125

Ignore, negative value

Thus, it takes 2 seconds for object to hit the ground

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