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dem82 [27]
2 years ago
10

What multiplication problem equals 160

Mathematics
2 answers:
rodikova [14]2 years ago
7 0
1*160
2*80
4*40
5*32
8*20
10*16 ..... Etc...
puteri [66]2 years ago
3 0

Answer:

Multiple possible answers

Step-by-step explanation:

1*160

2*80

4*40

5*32

8*20

10*16

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pishuonlain [190]

The domain is (-5,4). If that makes any sense, if it doesnt ill explain


4 0
3 years ago
Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane
liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

Answer:

Rate = 0.935042^\circ /cm

Step-by-step explanation:

Given

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

T(x,y) =x\sin2y

r = 1m

v = 2m/s

Express the given point P as a unit tangent vector:

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

u = \frac{\sqrt 3}{2}i - \frac{1}{2}j

Next, find the gradient of P and T using: \triangle T = \nabla T * u

Where

\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})}  = (sin \sqrt 3)i + (cos \sqrt 3)j

So: the gradient becomes:

\triangle T = \nabla T * u

\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] *  [\frac{\sqrt 3}{2}i - \frac{1}{2}j]

By vector multiplication, we have:

\triangle T = (sin \sqrt 3)*  \frac{\sqrt 3}{2} - (cos \sqrt 3)  \frac{1}{2}

\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)

\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5

\triangle T = 0.935042

Hence, the rate is:

Rate = \triangle T = 0.935042^\circ /cm

3 0
2 years ago
The sum of the angle measures of a polygon with n sides is 1260. find n.
Murrr4er [49]
Knowing the interior angle sum formula is; 

180x(n-2)

180x(n-2)=1260 (solve for n here) 
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180n=1260+360
180n=1620 
n=1620/180 
n=9

Therefore, this polygon has 9 sides.

Hope I helped :) 
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3 years ago
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Arada [10]

Answer:

187 students in 7th or 8th grade

17,088 pages

Step-by-step explanation:

412 - 225 is 187.

A dozen is 12

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