The ratio of their perimeters is still 7:20. The poster is a rectangle shape. If the length of each side of a poster is proportional to the other by a certain ratio, the sum of the lengths of all sides is also proportional by the same ratio. You can suppose the lengths of four sides of poster A are 7a, 7b, 7c, 7d, so poster B has 20a, 20b, 20c, 20d. (7a+7b+7c+7d)/(20a+20b+20c+20d)=7(a+b+c+d)/20(a+b+c+d)=7/20.
5x-5=-7y
3x+2y=-8
First, you need to rearrange the first equation so it is in the same format as the second one.
5x-5=-7y
Add 5 to both sides
5x-5+5=-7y+5
Add 7y to both sides
5x+7y=-7y+7y+5
So you have 5x+7y=5
Now you need to multiply that equation by 3
3(5x+7y=5)
15x+21y=15
Multiply the second equation by -5
-5(3x+2y=-8)
-15x-10y=40
Now add them together
15x+21y=15
+(-15x-10y=40)
---------------------
11y=55
y=5
Now plug y=5 into one of the original equations and solve for x.
3x+2(5)=-8
3x+10=-8
3x=-8-10
3x=-18
x=-6
To check the solution plug them both into the other equation:
5(-6)-5=-7(5)
-30-5=-35
-35=-35
It checks.
Hope that helps.
Answer:
-ln|x−5| + 2 ln(x²+4) + 3 tan⁻¹(x/2) + C
Step-by-step explanation:
The fraction will be split into a sum of two other fractions.
The first fraction will have a denominator of x − 5. The numerator will the a polynomial of one less order, in this case, a constant A.
The second fraction will have a denominator of x² + 4. The numerator will be Bx + C.
Combine the two fractions back into one using the common denominator.
This numerator will equal the original numerator.
Match the coefficients.
Solve the system of equations.
So we can rewrite the integral as:
Solving:
$1.79 i hope this helps you
The answer is 3.12 thank me later ;)
