The answer of given equation is
+ 5c +4
Step-by-step explanation:
By putting the given values of E and F we get,
E + F
=(
) + (-
)
Here we will take the coefficient of same variable together
=(
) +( -2c +7c) +( -1+5)
=
+ 5c +4
Hope this helps you.
YoU sHoUlD pRoBaBlY cHiLl aNd ThInK aBoUt A uNiCoRn eAtInG a TrUcKlOaD oF cHeEzE aT cHiLi’S.
P = 2(L + W)
P = 114
W = 23
114 = 2(L + 23)
114 = 2L + 46
114 - 46 = 2L
68 = 2L
68/2 = L
34 = L <=== so the length is 34 ft
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:
11.53 inches
Step-by-step explanation:
Given data
Diagonal = 19 inches
Height= 15.1 inches
Let us apply the Pythagoras theorem
D^2= H^2+W^2
substitute
19^2= 15.1^2+W^2
361=228.01+W^2
361-228.01=W^2
132.99= W^2
W=√132.99
W=11.53 inches
Hence the width is 11.53 inches