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2 years ago

Find the surface area of the prism. 16cm 12cm 10cm 20cm

Mathematics

1 answer:

kykrilka [37]2 years ago

5 0

Answer:

Step-by-step explanation:

<h3>I hope it's helpful for you</h3>

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Please help!!! I’ll mark you as brainliest!!!!

Daniel [21]

Answer: D $26

Step-by-step explanation:

50-(50x.3)= 35

35-6=29

29-3=26

5 0

3 years ago

Find the side length. Round to the nearest tenth if necessary.<br> 15

Rasek [7]

Answer:

x = 39

Step-by-step explanation:

This is pythagoras

x = √36² + 15²

x = √1296 + 225

x = √1521

x = 39

Hopefully it helps you

Ask if u don't understand

8 0

3 years ago

Help please I do not know how to do this

sleet_krkn [62]

Answer:

0 and 1

Step-by-step explanation:

If a polynomial P(x) is divided by (x + h), then by the Remainder theorem

The remainder is the value of P(- h)

(1)

For P(x) divided by (x - 1) , then the remainder is P(1) = 0

(2)

For P(x) divided by (x + 2) then the remainder is P(- 2) = 1

6 0

3 years ago

Evaluate and simplify the following complex fraction.

likoan [24]

Answer:

yjyjytjyjtyjyjtyjtyjyhjyjnfrhjnrthj

Step-by-step explanation:

8 0

2 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