1.1x+1.2x-5.4=-10
11/10x+12/10x-54/10=10
11x/2.5+2^2-1*3/5x-3^3/5=-10
(11x)+2(2*3x)+2(-3^3)/2*5=-10
11x+2(6x)+2(-27)/2*5=-10
11x+2*6x-2*27/2*5=-10
11x+12x-54/2*5=-10
23x-54/2*5=-10
23x-54=-100
23x=-46
23x/23=-46/23
x=-2*23/23
x=-2
Answer:
75
Step-by-step explanation:
f(1) = 7
f(n) = 3f(n-1) + 3
So what you are trying to do here is find the recursive value (that's what it is called) for f(n). Computers love this sort of thing, but we humans have to do it slowly and carefully.
So let's try f(2)
That means that f(2) = 3*f(n-1) + 3
but if f(2) is used it means that you have to know f(2-1) which is just f(1) and we know that.
so f(2) = 3*f(1)+3
f(2) = 3*7 + 3
f(2) = 21 + 3
f(2) = 24
Now do it again. We now know F(2), so we should be able to find f(3)
f(3) = 3*f(3 - 1) + 3
f(3) = 3*f(2) + 3 We know that f(2) = 24
f(3) = 3* 24 + 3
f(3) = 72 + 3
f(3) = 75
Answer:
c
Step-by-step explanation:
Here's how this works:
Get everything together into one fraction by finding the LCD and doing the math. The LCD is sin(x) cos(x). Multiplying that in to each term looks like this:
![[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?](https://tex.z-dn.net/?f=%5Bsin%28x%29cos%28x%29%5D%5Cfrac%7Bsin%28x%29%7D%7Bcos%28x%29%7D%2B%5Bsin%28x%29cos%28x%29%5D%5Cfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%20%3D%3F)
In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:
Put everything over the common denominator now:
Since 
, we will make that substitution:
We could separate that fraction into 2:
×
and 
Therefore, the simplification is
sec(x)csc(x)
