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2 years ago

What is the area of this figure?

Mathematics

1 answer:

Dima020 [189]2 years ago

4 0

Answer:

$220.5 \: {yd}^{2}$

Step-by-step explanation:

Area of the figure = Area of the arc with radius 10 yd and central angle 90° + Area of rectangle with dimensions (10 + 5 - 3 = 12) 12 yd and (7 + 6 - 4 = 9) 9 yd + Area of square with dimension 4yd + Area of rectangle with dimensions 3 yd by 2 yd + Area of triangle with base 3 yd and height (5 + 3 = 8) 8 yd.

$= \frac{90 \degree}{360 \degree} \times 3.14 {(10)}^{2} + 12 \times 9 + {4}^{2} \\ + 3 \times 2 + \frac{1}{2} \times 3 \times 8 \\ \\ = \frac{1}{4} \times 314 + 108 + 16 + 6 + 12 \\ \\ = 78.5 + 142 \\ \\ = 220.5 \: {yd}^{2}$

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Rama09 [41]

Question:

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?

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Answer:

It takes 2 seconds for object to hit the ground

Solution:

The given equation is:

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

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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2 years ago

Which pair of triangles can be proved congruent by the SAS postulate?

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