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2 years ago

15

Solve the equation for the indicated variable.1)7r+1=10s for "r"

1 answer:

Oksana_A [137]2 years ago

4 0

Given:

The equation is given as,

$7r+1=10s$

The objective is to solve the equation in terms of <em>r</em>.

Explanation:

The equation can be simplified as,

$\begin{gathered} 7r+1=10s \\ 7r=10s-1 \\ r=\frac{10s-1}{7} \end{gathered}$

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During a 90-day monsoon season, daily rainfall can be modeled by sinusoidal functions. A low of 4 inches of rainfall was recorde

Phoenix [80]

Answer:

$25.4\,days \leq t \leq 34.6\,days$

Step-by-step explanation:

Let consider the following model:

$c(t) = A \cdot \sin \left(\frac{2\pi \cdot t}{T} \right) + \bar c$

The average is given by the following formula:

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The maximum value is:

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$c_{max} = 12\,in$

Amplitude is:

$A = 12\,in - 8\,in$

$A = 4\,in$

The sine function has a periodicity of $2\pi$ , where is minimum is reached at $\theta = \frac{3\pi}{2}$ , when t = 30. The period of the cycle is:

$T = \frac{2\pi}{\frac{3}{2}\pi }\cdot (30)$

$T = 40$

The complete expression is:

$c(t) = 8\,in + 4\,in \cdot \sin \left(\frac{2\pi}{40}\cdot t \right)$

The times associated with $c = 5\,in$ are, respectively:

$8\,in + 4\,in \cdot \sin \left(\frac{2\pi}{40}\cdot t \right) = 5\,in$

$4\,in \cdot \sin \left(\frac{2\pi}{40}\cdot t \right) = -3\,in$

$\sin \left(\frac{2\pi}{40}\cdot t \right) = -0.75$

$t = \frac{40}{2\pi}\cdot \sin^{-1} (-0.75)$

Instants are, respectively:

$t_{1} \approx 25.4\,days$

$t_{2} \approx 34.6\,days$

Period is:

$25.4\,days \leq t \leq 34.6\,days$

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3 years ago

A sphere has a diameter of 15inches. what is the volume of the sphere rounded to the nearest tenth? use 3.14 for pi. A:441.6in 3

algol13

<u>Answer:</u> B.

Step-by-step explanation:

Diameter = Radius divided by 2.

The volume formula of a sphere is V = 4/3 π r^3.

Radius of this problem = 7.5.

So, V = 4/3 π (7.5) ^3.

Once you "plug in" the radius and solve the equation, you get an answer of 1,766.3, rounded to the nearest tenth.

<u>Don't forget the inches cubed!</u>

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