Answer:
$9,220,000(0.888)^t
Step-by-step explanation:
Model this using the following formula:
Value = (Present Value)*(1 - rate of decay)^(number of years)
Here, Value after t years = $9,220,000(1 -0.112)^t
Value after t years = $9,220,000(0.888)^t
Remark
There is no short way to do this problem and no obvious way to get the answer other that to solve each part.
Solve
A
Multiply by 2
x + 1.6 = 2(x + 0.1) Remove the brackets
x + 1.6 = 2x + 0.1*2
x + 1.6 = 2x + 0.2 Subtract x from both sides
1.6 = x + 0.2 Subtract 0.2 from both sides
1.6 - 0.2 = x
1.4 = x
Circle A
B
Subtract 2x from both sides.
3x - 2x = 1.4
Circle B
C
Remove the brackets.
4x + 6 = 2x - 6 Add 6 to both sides
4x + 12 = 2x Subtract 4x from both sides.
12 = -2x Divide by - 2
12/-2 = x
x = - 6 Don't circle C
D
I'm going to be very scant in my solution of this. You can fill in the steps.
3x = 4.2
x = 4.2/3
x = 1.4
Circle D
Harap awak faham hehehe kalau jawapan betul bagi brainliest yer hehehe
Answer:
It diverges.
Step-by-step explanation:
We are given the inetegral: ![\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%5Cinfty%7D_2%20%5Cfrac%7B1%7D%7Bx%7D%20%28%5Cln%20x%29%5E2%20dx)
![\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx=\int\limits^{\infty}_2 (\ln x)^2 d(\ln x)=\\\\=\lim_{t \to \infty} \int\limits^t_2 (\ln x)^2d(\ln x)=\lim_{t \to \infty} \frac{(\ln t)^3}{3} |^t_2=\infty-\frac{(\ln 2)^3}{3} =\infty](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%5Cinfty%7D_2%20%5Cfrac%7B1%7D%7Bx%7D%20%28%5Cln%20x%29%5E2%20dx%3D%5Cint%5Climits%5E%7B%5Cinfty%7D_2%20%28%5Cln%20x%29%5E2%20d%28%5Cln%20x%29%3D%5C%5C%5C%5C%3D%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20%5Cint%5Climits%5Et_2%20%28%5Cln%20x%29%5E2d%28%5Cln%20x%29%3D%5Clim_%7Bt%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%28%5Cln%20t%29%5E3%7D%7B3%7D%20%7C%5Et_2%3D%5Cinfty-%5Cfrac%7B%28%5Cln%202%29%5E3%7D%7B3%7D%20%3D%5Cinfty)
So it is divergent.
-16 + (-8) = -24
(-4) - 7 = -11
-24 - (-9) = -15
6.25 + (-8.50) = 2.25
10.8 - (6.4) = 4.4
-54.26 + (-15.42) = -69.68
F ur fractions