<h2>You need a reflection across the y-axis!</h2>
Answer:
m<A = 107
m<B = 73
m<C = 107
m<D = 73
Step-by-step explanation:
2(4x + 1) + 2(5x + 17) = 360
8x + 2 + 10x + 34 = 360
18x + 36 = 360
18x = 324
x = 18
4x + 1
4(18) + 1
73
5x + 17
5(18) + 17
107
Answer:
15 feet
Step-by-step explanation:
the perimeter is the total length of the outside of a shape. 2+3+10=15
Let P be Brandon's starting point and Q be the point directly across the river from P.
<span>Now let R be the point where Brandon swims to on the opposite shore, and let </span>
<span>QR = x. Then he will swim a distance of sqrt(50^2 + x^2) meters and then run </span>
<span>a distance of (300 - x) meters. Since time = distance/speed, the time of travel T is </span>
<span>T = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x). Now differentiate with respect to x: </span>
<span>dT/dx = (1/4)*(2500 + x^2)^(-1/2) *(2x) - (1/6). Now to find the critical points set </span>
<span>dT/dx = 0, which will be the case when </span>
<span>(x/2) / sqrt(2500 + x^2) = 1/6 ----> </span>
<span>3x = sqrt(2500 + x^2) ----> </span>
<span>9x^2 = 2500 + x^2 ----> 8x^2 = 2500 ---> x^2 = 625/2 ---> x = (25/2)*sqrt(2) m, </span>
<span>which is about 17.7 m downstream from Q. </span>
<span>Now d/dx(dT/dx) = 1250/(2500 + x^2) > 0 for x = 17.7, so by the second derivative </span>
<span>test the time of travel, T, is minimized at x = (25/2)*sqrt(2) m. So to find the </span>
<span>minimum travel time just plug this value of x into to equation for T: </span>
<span>T(x) = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x) ----> </span>
<span>T((25/2)*sqrt(2)) = (1/2)*(sqrt(2500 + (625/2)) + (1/6)*(300 - (25/2)*sqrt(2)) = 73.57 s.</span><span>
</span><span>
</span><span>
</span><span>
</span><span>mind blown</span>
Answer:
4
Step-by-step explanation:
![\huge \sqrt[4]{256} = \sqrt[4]{ {4}^{4} } = 4](https://tex.z-dn.net/?f=%20%5Chuge%20%5Csqrt%5B4%5D%7B256%7D%20%20%3D%20%20%5Csqrt%5B4%5D%7B%20%7B4%7D%5E%7B4%7D%20%7D%20%3D%204%20%20)
