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2 years ago

What is .61 repeating as a fraction and how did you get the answer

1 answer:

7 0

61/100, because if it is 0.(value) than it is value / 100 assuming value is 2 digits long. We cannot simplify it further because 61 is prime number.

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Please help !!!!picture shown

dexar [7]

<span>A. H(w) gets very large. </span>

7 0

3 years ago

The function h (d) = 2d + 4.3 relates the height (h) of the water in a fountain in feet to the diameter (d) of the pipe carrying

mezya [45]

Answer:

h(1.5) = 7.3 ft

h(10.3) = 24.9 ft

Step-by-step explanation:

Given the function h(d) = 2d + 4.3,

where:

h = height of the water in a fountain (in feet)

d = diameter of the pipe carrying the water (in inches)

Substitute the input value of d = 1.5, into the function:

h(1.5) = 2(1.5) + 4.3

h(1.5) = 3 + 4.3

h(1.5) = 7 feet

The height of the water in a fountain is 7 feet when the diameter of the pipe is 1.5 inches.

Substitute the input value of d = 10.3, into the function:

h(10.3) = 2(10.3) + 4.3

h(10.3) = 20.6 + 4.3

h(10.3) = 24.9 feet

The height of the water in a fountain is 24.9 feet when the diameter of the pipe is 10.3 inches.

<h3>Context of the solutions to h(1.5) and h(10.3):</h3>

The solutions to both functions show the relationship between the diameter of the pipe to the height of the water in a fountain. The height of the water in fountain increases relative to the diameter of the pipe. In other words, as the diameter or the size of the pipe increases or widens, the height of the water in a fountain also increases.

3 0

2 years ago

Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin

saw5 [17]

Treat $\mathcal C$ as the boundary of the region $\mathcal S$ , where $\mathcal S$ is the part of the surface $z=2xy$ bounded by $\mathcal C$ . We write

$\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r$

with $\mathbf f=(y+7\sin x,z^2+9\cos y,x^3)$ .

By Stoke's theorem, the line integral is equivalent to the surface integral over $\mathcal S$ of the curl of $\mathbf f$ . We have

$\nabla\times\mathbf f=(-2z,-3x^2,-1)$

so the line integral is equivalent to

$\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$
$=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\mathbf s(u,v)$ is a vector-valued function that parameterizes $\mathcal S$ . In this case, we can take

$\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)$

with $0\le u\le1$ and $0\le v\le2\pi$ . Then

$\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$

and the integral becomes

$\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi$ <span />

4 0

2 years ago

find the length of the segment connecting the two given points. Write your answer in simplest form . ** Round your answer to the

taurus [48]

Answer: 16

Step-by-step explanation:

Because the Y is 0 for both, you can simply draw a line graph with one point at 15, and the other at -1. The difference between 15 and -1 is 16. Or ABS(15-(-1))=16

8 0

2 years ago

What measure of central tendency is most affected by outliers?.

Elden [556K]

Answer: Mean

Mean is the only measure of central tendency that is always affected by an outlier.

hope this helps!

have a nice day!

8 0

2 years ago