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yuradex [85]
2 years ago
10

A. figure P B. figure R C. figure S D. figure G

Mathematics
1 answer:
vodomira [7]2 years ago
4 0

Answer:

D. figure G

Step-by-step explanation:

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Polygons QRST and Q′R′S′T′ are shown on the following coordinate grid:
Rudik [331]
ANSWER : QR is the answer
3 0
3 years ago
Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap
pantera1 [17]

Answer:

The height of the pile is increasing at the rate of  \mathbf{ \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e \dfrac{dV}{dt}= 20 \ ft^3/min

we know that radius r is always twice the   diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

V = \dfrac{\pi r^2 h}{3}

V = \dfrac{\pi (h/2)^2 h}{3}

V = \dfrac{1}{12} \pi h^3

V = \dfrac{ \pi h^3}{12}

Taking the differentiation of volume V with respect to time t; we have:

\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}

\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}

we know that:

\dfrac{dV}{dt}= 20 \ ft^3/min

So;we have:

20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}

20= 56.25 \pi \times \dfrac{dh}{dt}

\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

The height of the pile is increasing at the rate of  \mathbf{ \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

8 0
3 years ago
Please help with the question in the picture!
Stella [2.4K]

Answer:

Tan C = 3/4

Step-by-step explanation:

Given-

∠ A = 90°, sin C = 3 / 5

<u>METHOD - I</u>

<u><em>Sin² C + Cos² C = 1</em></u>

Cos² C = 1 - Sin² C

Cos² C = 1 - \frac{9}{25}

Cos² C = \frac{25 - 9}{25}

Cos² C = \frac{16}{25}

Cos C = \sqrt{\frac{16}{25} }

Cos C = \frac{4}{5}

As we know that

Tan C = \frac{Sin C}{Cos C }

<em>Tan C = \frac{\frac{3}{5} }{\frac{4}{5} }</em>

<em>Tan C = \frac{3}{4}</em>

<u>METHOD - II</u>

Given Sin C = \frac{3}{5} = \frac{Height}{Hypotenuse}

therefore,  

AB ( Height ) = 3; BC ( Hypotenuse) = 5

<em>∵ ΔABC is Right triangle.</em>

<em>∴ By Pythagorean Theorem-</em>

<em>AB² + AC² = BC²</em>

<em>AC² </em><em>= </em><em>BC² </em><em>- </em><em> AB</em><em>² </em>

<em>AC² = 5² - 3²</em>

<em>AC² = 25 - 9</em>

<em>AC² = 16</em>

<em>AC  ( Base) = 4</em>

<em>Since, </em>

<em>Tan C = \frac{Height}{Base}</em>

<em>Tan C = \frac{AB}{AC}</em>

<em>Hence Tan C = \frac{3}{4}</em>

<em />

4 0
3 years ago
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Answer: D

Step-by-step explanation:

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Step-by-step explanation:

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