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jolli1 [7]
3 years ago
13

Solve for b. -11= -47 - 12h

Mathematics
1 answer:
Butoxors [25]3 years ago
7 0

Answer:

h = -3

Step-by-step explanation:

-11 = -47 - 12h

First you need to get "h" alone on one side.

To do that, you need to add 47 to both sides:

47-11 = -12h

36 = -12h

Next, you need to divide -12 by both sides, to get "h" alone:

36/-12 = -12h/-12

-3 = h

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george has to pay $30 in taxes for every $100 that he earns.Last summer he earned $3,680.How much did he pay in taxes.
Ludmilka [50]
We can solve this problem by rule of 3:

$30 (taxes)------------$100
x-----------------------$3,680

x=($30*$3,680) / $100=$1,104


Answer: $1,104 paid  in taxes.
7 0
4 years ago
The position of an object attached to a spring is given by y(t) = cos(5t) − sin(5t), where t is time in seconds. In the first 4
fredd [130]

Answer:

  6

Step-by-step explanation:

The object reaches an extreme position 6 times in the first 4 seconds. At each of those positions, the velocity is zero (as it changes sign).

5 0
3 years ago
The area of a semicircle is 2 times the area of a circle.<br><br> True<br><br> False
Brrunno [24]
The Area of a semicircle is half of the area of a circle.

In short, Your statement would be: "False"

Hope this helps!
3 0
3 years ago
4736 divided by 100 for math
damaskus [11]
If you divided 4.736 ÷ 100 you have to find the times 100 is in 473, is 4 times, then you have to find how many times is 100 in 736, is 7 times, then you have 36 but this is not a entire number then you add a 0, 100 is 3 times in 360, now you have a 6, you add two zeros, 100 is six times in 600. 47,36 is the answer.
6 0
3 years ago
A particle moves along line segments from the origin to the points (2, 0, 0), (2, 3, 1), (0, 3, 1), and back to the origin under
sweet [91]

Answer: Work done by the particle is 78 N.

Step-by-step explanation:

If C is the path the particle follows, then work done is \int_{C}^{}F.dx.

According to the question the force F=z^{2}i+5xyj+4y^2k and all four points are in the plane z=\frac{1}{4}y.

therefore, if S is the at surface with boundary C, so that S is the portion of the plane z=\frac{1}{4}y over the rectangle D=[0,2]\times [0,4].

Now,

curlF=\begin{vmatrix}i &j  &k \\  \frac{\partial }{\partial x}&\frac{\partial }{\partial y}  &\frac{\partial }{\partial z} \\  z^{2}&5xy  &4y^{2} \end{vmatrix}=8yi+2zj+5yk

By the Stoke's theorem:

\int_{C}^{}F.dx=\iint_{s}^{}curlF.dS\\\\=\iint_{D}^{}[-8y(0)-2z\left ( \frac{1}{4} \right )+5y]dA\\\\=\iint_{D}^{}\left ( -\frac{1}{2}z+5y \right )dA

=\iint_{D}^{}\left ( \frac{39}{8}y \right )dA\, \, \, \, \, \, \, \, \, Since\, \, z=\frac{1}{4}y\\\\=\int_{0}^{2}\int_{0}^{4}\frac{39}{8}y\, \, \, dy\, dx\\\\

\int_{0}^{2}\left [ \frac{39}{16}y^2 \right ]_{0}^{4}dx\\\\=\int_{0}^{2}39\, \,  dx\\\\=39\times 2\\\\=78 \, N

8 0
4 years ago
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