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Effectus [21]
3 years ago
5

9 percent to decimal to thousands place

Mathematics
1 answer:
kiruha [24]3 years ago
7 0

Answer:

0.009 is a decimal and 0.9/100 or 0.9% is the percentage for 9/1000.

Step-by-step explanation:

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A man travels 320 miles in 8 hours. If he continues at the same rate, how many miles will he
evablogger [386]

Answer: 80 miles

Step-by-step explanation:

320miles/8hours= 40mph

40mph=xmiles/2hrs

40*2= 80 miles

8 0
3 years ago
How would you make 0.24 as a mixed number
MrRissso [65]

0.24 = 24/100 = 6/25

3 0
4 years ago
What is the radius of a circle with an area of 50.24 cubic inches?<br><br> Use 3.14 for pi.
OverLord2011 [107]
Hope this helps you..!!!

3 0
4 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7Br%7D%7B2%7D%2B%5Cfrac%7Bs%7D%7Br%7D%20-%5Cfrac%7Br%7D%7B4%7D%2B%5Cfrac%7B1%7D%7B5%7D
zloy xaker [14]

Answer:

-1/20

Step-by-step explanation:

r/2 + s/r - r/4 + 1/5 =

= -1/2 + 0/(-1) - (-1)/4 + 1/5

= -1/2 + 1/4 + 1/5

= -10/20 + 5/20 + 4/20

= -5/20 + 4/20

= -1/20

3 0
3 years ago
Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
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