Who know how to do this ?

What is the approximate area of the shaded portion of the diagram? (Use 3.14 as an estimate for pi.)

Nadusha1986 [10]

Answer:

218

Step-by-step explanation:

first find the area of the circle-

$A=\pi r^{2} \\A=\pi 10^{2} \\A=100\pi \\A=314$

to find the area of the triangle, we need the base and the height. The base we know is 16, since we don't know the height yet, use the Pythagorean theorem to find the height. One leg is 16 the hypotenuse is 20 (2* the radius).

$a^{2} +b^{2} =c^{2} \\a^{2} +16^{2} =20^{2} \\a^{2} +256=400\\a^{2} =144\\a=12$

Now I know the height is 12. Find the area of the triangle

$A=\frac{1}{2} bh\\A=\frac{1}{2} (16)(20)\\A=96$

subtract the area of the triangle from the area of the circle

314-96=218

6 0

2 years ago

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 73 and 9

Naddika [18.5K]

Answer: After about 9.03 hours the temperature first reach 82 degrees.

Step-by-step explanation:

The sinusoidal function is given by :

$y=A\sin[\omega(x-\alpha)]+C$

where, A = amplitude; $\omega=\dfrac{2\pi}{period}$ , α= phase shift on the Y-axis and C = midline.

As per given,

Average daily temperature= $C=\dfrac{73+97}{2}=85$ [midline is average of upper and lower limit.]

A= 97-85 = 12

Phase shift: $\alpha=10$

Period = 24 hours;

$\omega=\dfrac{2\pi}{24}=\dfrac{\pi}{12}$

Substitute all values in sinusoidal function, we get

$y=12\sin[\dfrac{\pi}{12}(x-10)]+85$

Put y= 82, we get

$82=12\sin[\dfrac{\pi}{12}(x-10)]+85\\\\\Rightarrow\ -3= 12\sin[\dfrac{\pi}{12}(x-10)]\\\\=\dfrac{-1}{4}= \sin[\dfrac{\pi}{12}(x-10)]\\\\\Rightarrow\ \dfrac{\pi}{12}(x-10)=\sin^{-1}(\dfrac{-1}{4})\\\\\Rightarrow\ x-10=\dfrac{12}{\pi}(\sin^{-1}(\dfrac{-1}{4}))\\\\\Rightarrow\ x=\dfrac{12}{\pi}(\sin^{-1}(\dfrac{-1}{4}))+10\\\Rightarrow\ x\approx9.03$