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mylen [45]
3 years ago
5

HELPP !!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Umnica [9.8K]3 years ago
3 0

I cannot see the picture to good
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Are there any numbers that go into 78 and 780
Kryger [21]
Are there any numbers that go into 78 and 780?

78 ÷ 2 = 39
780 ÷ 2 = 390

2 can be a number that goes in 78 and 780.
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3 years ago
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100 Points!!! I will give Brainliest, a 5 star rating and a thank you. pls help me with question 1 through 5. 1 and 2 is on atta
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1. A. similar

2. C.

3. B. all congruent figures are also similar figures.

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$500 shared in ratio 2:3:9
RideAnS [48]
2:3:9......added together = 14

2/14 (500) = 1000/14 = 71.43
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9/14 (500) = 4500/14 = 321.43
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Please help me I’m working on a math test and I’m struggling with this problem: The Aon center in Chicago is one of the worlds t
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4 0
2 years ago
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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
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