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3 years ago

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The annual snowfall in city A is 58.7 inches. This is 12.5 inches more than six times the snowfall in city B. Find the annual sn

owfall for city B.

2 answers:

777dan777 [17]3 years ago

4 0

Since there is 6 times more snow in city B than in city A, 6 times 58.7 is 352.2.

Now, the problem say says, " <span>This is 12.5 inches more than six times the snowfall in city B" so, you have to add 12.5 to 352.2.

352.2+12.5= 364.7

ANSWER: The annual snowfall in city B is 364.7 inches.
</span>

Lelechka [254]3 years ago

3 0

The annual snowfall for city B is 7.7 inches.
Since it is 12.5 inches more: 58.7 - 12.5 = 46.2
Since 46.2 is 6 times that of city B: 46.2 / 6 = 7.7

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Line segment NY has endpoints N(-11, 5) and Y(3,-3).

777dan777 [17]

Given:

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

To find:

The equation of the perpendicular bisector of NY.

Solution:

Midpoint point of NY is

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Midpoint=\left(\dfrac{-11+3}{2},\dfrac{5-3}{2}\right)$

$Midpoint=\left(\dfrac{-8}{2},\dfrac{2}{2}\right)$

$Midpoint=\left(-4,1\right)$

Slope of lines NY is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m=\dfrac{-3-5}{3-(-11)}$

$m=\dfrac{-8}{14}$

$m=\dfrac{-4}{7}$

Product of slopes of two perpendicular lines is -1. So,

$m_1\times \dfrac{-4}{7}=-1$

$m_1=\dfrac{7}{4}$

The perpendicular bisector of NY passes through (-4,1) with slope $\dfrac{7}{4}$ . So, the equation of perpendicular bisector of NY is

$y-y_1=m_1(x-x_1)$

$y-1=\dfrac{7}{4}(x-(-4))$

$y-1=\dfrac{7}{4}(x+4)$

$y-1=\dfrac{7}{4}x+7$

Add 1 on both sides.

$y=\dfrac{7}{4}x+8$

Therefore, the equation of perpendicular bisector of NY is $y=\dfrac{7}{4}x+8$ .

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3 years ago

A diver starts at the surface of the water and travels 8 feet to the bottom.

Sauron [17]

Answer:

On the graph B, at 0 seconds the graph will be <u>at $y=0.$ </u>

The graph will be <u>going down</u> until 2 seconds, when the diver reaches her deepest point. At 2 seconds the height of the graph will be -8ft.

Step-by-step explanation:

In graph B we are measuring the distance from the surface, that is we are setting the surface to be y=0. Thus if the diver reaches her deepest point 8ft down, she will be below y=0 and at -8ft.

Thus, in shape the graph B will be similar to graph A, but it will be shifted downed by 8ft.

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4 years ago

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3 years ago

One car alarm is set to go off every 10 seconds. Another car alarm is set to go off every 5 seconds, and a third car alarm is se

erma4kov [3.2K]

Answer:

20 seconds

Step-by-step explanation:

Car A = every 10 seconds

Car B = every 5 seconds

Car C = every 4 seconds

If they all start to go off at the same time due to a loud clap of thunder, how long will it take until they all sound at the same time again?

To determine how long will it take until they all sound at the same time again, find the lowest common multiples of cars alarm time

Car A (every 10 seconds) = 20, 30, 40, 50, 60, 70, 80, 90

Car B (every 5 seconds) = 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

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Therefore, it will take 20 seconds for they all sound at the same time again

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3 years ago

For the following integral, find the approximate value of the integral with 4 subdivisions using midpoint, trapezoid, and Simpso

PIT_PIT [208]

Answer:

$\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}$

$\textsf{Trapezium rule}: \quad \pi$

$\textsf{Simpson's rule}: \quad \dfrac{4 \pi}{3}$

Step-by-step explanation:

<u>Midpoint rule</u>

$\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

<u>Trapezium rule</u>

$\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}$

<u>Simpson's rule</u>

$\displaystyle \int_{a}^{b} y \:\:\text{d}x \approx \dfrac{1}{3}h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{n-2}+4y_{n-1}+y_n\right)\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

<u>Given definite integral</u>:

$\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x$

Therefore:

a = 0
b = 2π

Calculate the subdivisions:

$\implies h=\dfrac{2 \pi - 0}{4}=\dfrac{1}{2}\pi$

<u>Midpoint rule</u>

Sub-intervals are:

$\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]$

The midpoints of these sub-intervals are:

$\dfrac{1}{4} \pi, \dfrac{3}{4} \pi, \dfrac{5}{4} \pi, \dfrac{7}{4} \pi$

Therefore:

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}$

<u>Trapezium rule</u>

$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} &&&&&\\ x & 0 & \dfrac{1}{2}\pi & \pi & \dfrac{3}{2} \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}$

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}$

<u>Simpson's rule</u>

<u />

<u /> $\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}$

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2 years ago