Hello good person. The answer is C, (4,7). If you replace the x's in both equations (4) and the y's i both equations (7) you would get your answer for both.

4 0

3 years ago

the circle below has center T. suppose that m UV = 138 degrees and that UW is tangent to the circle at U. find the following

alisha [4.7K]

Angle UTV is basically given: it's 138°.

Moreover, since UTV is isosceles, the other two angles have the same measure x. We thus have

$138+x+x=180 \iff 2x=42 \iff x=21$

So, TUV is 21°, and VUW is complementary, so it must be 69° so that they sum up to 90°

3 0

3 years ago

The rate of transmission in a telegraph cable is observed to be proportional to x2ln(1/x) where x is the ratio of the radius of

sergij07 [2.7K]

Answer:

The value of x that gives the maximum transmission is 1/√e ≅0.607

Step-by-step explanation:

Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.

$f'(x) = k*((x^2)'*ln(1/x) + x^2*(ln(1/x)')) = k*(2x\,ln(1/x)+x^2*(\frac{1}{1/x}*(-\frac{1}{x^2})))\\= k * (2x \, ln(1/x)-x)$

We need to equalize f' to 0

k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
ln(1/x) = x/2x = 1/2 ------- we send the natural logarithm as exp
1/x = e^(1/2)
x = 1/e^(1/2) = 1/√e ≅ 0.607

Thus, the value of x that gives the maximum transmission is 1/√e.

7 0

3 years ago