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3 years ago

8

Which function represents the graph of h(x)=2∣∣x+3∣∣−1h(x)=2|x+3|−1 after it is translated 2 units right?

1 answer:

frosja888 [35]3 years ago

6 0

C because i learned it from xxxx xxx

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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

Please help: Triangle ABC is similar to triangle PQR where

andre [41]

Explanation:

AB=7
AC=9
You are looking for R

7+9=16

So R is 16

6 0

2 years ago

Can you help me please

stellarik [79]

Answer:

sine 23 = x/21

Multiply both sides by 21

21 x sine 23 = x

Now, use a calculator and find the answer of the 21 times the sine of 23:

x = -17.7706284876

8 0

2 years ago

Point A(3,2) is on a circle whose center is C(−2,3). What is the radius of the circle?

goldfiish [28.3K]

Answer:

The radius of the circle is √26.

Step-by-step explanation:

Since point A is on the circle, the distance between that point and the center of the circle is equal to the radius.

To find the distance between two points, the following equation should be used:

$x^{2} =(x_{A}- x_{C}) ^{2} + (y_{A}- y_{C}) ^{2}$

Replacing the numeric values:

$r^{2} =(3- (-2)) ^{2} + (2-3) ^{2}\\r^{2} =(5) ^{2} + (-1) ^{2}\\r=\sqrt{26}$

The radius of the circle is √26.

6 0

3 years ago

Please help me now I need the answer to this now please help me please help

Vadim26 [7]

Answer:

Too much please's

Step-by-step explanation:

Please stop begging people to help.

3 0

2 years ago

Read 2 more answers