Answer: x=2 - 1/2 x ㏒(16/2953125)
Step-by-step explanation:
- log (28^2) + log(525^-3) + 2x=log(10^4)
- log(28^2 x 1/523^3) + 2x=4
- log(28^2/525^3) + 2x=4
- log (112/75 x 75 x 525) + 2x=4
- log(16/2953125) + 2x=4
- 2x=4-log(16/2953125)
- <u><em>x=2-1/2 x log(16/2953125)</em></u>
To find the slope you use a formula like y=mx+b y would be the point of whatever point is on the y axis so for example lets say 6 is your y axis and 8 is your a axis point once you have those point then you graph them. after they have been graphed you find the next point that would go in the sequence which would be at -8,-6 draw a line connecting the two point and then on every corner that the line completely goes through those are your other points then from there you take your original point and the next point closets to the original one and count down and to the left or right how ever many times you need to until you reach the nearest point. if you cant figure it out them this well help www.desmos.com/calculator
Answer:
x=15º
Step-by-step explanation:
A triangle's angles add up to 180º.
∠A=30º
∠C=180º-45º
∠B=xº
To find ∠B we have to first find ∠C.
Since ∠C does not have an angle, knowing that a straight line is 180º we can subtract it from 45º to find ∠C.
180-45=135
∠C=135º
Now add ∠A and ∠C.
30+135=165
Subtract the sum from 180º
180-165=x
x=15º
Hope this helps :)
Answer
I am not 100% sure on this, but I think that this is the answer. x=8
Step-by-step explanation:
Here is the equation that I made
4x+17+x+23=80
Let me know if it was right :)
Answer:
5.09 units
Step-by-step explanation:
Given equation
in the interval 
So we integrate 
in the given interaval
Let us integrate 
first.
let
Using integration by parts we get
So here
![\int f(x)=2\int\limits^4_1 {\ln x}dx\\ =2(x\ln x-x)_1^4\\ =2[(4\ln 4-4)-(1\ln 1-1)]\\ =2[4\ln 4-4+1]\\ =5.09\ units](https://tex.z-dn.net/?f=%5Cint%20f%28x%29%3D2%5Cint%5Climits%5E4_1%20%7B%5Cln%20x%7Ddx%5C%5C%20%3D2%28x%5Cln%20x-x%29_1%5E4%5C%5C%20%3D2%5B%284%5Cln%204-4%29-%281%5Cln%201-1%29%5D%5C%5C%20%3D2%5B4%5Cln%204-4%2B1%5D%5C%5C%20%3D5.09%5C%20units)
The area of the the region between the curve and horizontal axis is 5.09 units.
