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3 years ago

Write a story about temperatures that this expression could represent: 27 + (-11)

Mathematics

1 answer:

xxTIMURxx [149]3 years ago

7 0

Answer:

at noon the temperature was 27 degrees. At night, the sun went down and the temperature decreased by 11 degrees.

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Please help I really need it on this one thank you

Bond [772]

(-7) * 8 = -56

answer
A. -56

hope it helps

7 0

3 years ago

Phân tích thành thừa số <br>a) 2√46 + √69<br>b) x√x - y√y<br>c) x-2+√x^2-4

const2013 [10]

Need help to with that qustion

4 0

3 years ago

A family spent $93 at a carnival. They spent $18 on tickets to the carnival and $36 on food.They spent the rest of the money on

k0ka [10]

Answer:

$93-$18-$36=$39

$39

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A car begins to depreciate at a rate of 24.9% annually as soon as it is driven off the lot. If a car was purchased for 26,500; h

STatiana [176]

Answer:

13197

Step-by-step explanation:

You have to do 24.9*2 and then find that percent of 26500

6 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