4/5
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Hello here is a solution :
f'(x) =( sin11x)'cos11x +<span>sin11x(cos11x)'
(sin(ax+b))'=a cos(ax+b)
</span>(cos(ax+b))'=-a sin(ax+b)
f'(x) = 11 cos11xcos11x-11sin11xsin11x
f'(x) = 11(cos11x)²-11(sin11x)²
Answer:
Step-by-step explanation:
Notice that you are given an isosceles right-angle triangle to solve, since each of its two acute angles measures 
. Then such means that the sides opposite to these acute angles (the so called "legs" of this right angle triangle) must also be of the same length (x).
We can then use the Pythagorean theorem that relates the square of the hypotenuse to the addition of the squares of the triangles legs:
We use just the positive root, since we are looking for an actual length. then, the requested side is:
3(x-1)-8=4(1+x)+5
One solution was found :
x = -20
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
3*(x-1)-8-(4*(1+x)+5)=0
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((3•(x-1))-8)-(4•(x+1)+5) = 0
Step 2 :
Equation at the end of step 2 :
(3 • (x - 1) - 8) - (4x + 9) = 0
Step 3 :
Step 4 :
Pulling out like terms :
4.1 Pull out like factors :
-x - 20 = -1 • (x + 20)
Equation at the end of step 4 :
-x - 20 = 0
Step 5 :
Solving a Single Variable Equation :
5.1 Solve : -x-20 = 0
Add 20 to both sides of the equation :
-x = 20
Multiply both sides of the equation by (-1) : x = -20
One solution was found :
x = -20
hope this is wht u wanted
The two numbers are 14 and 15.
