a = interest rate of first CD
b = interest rate of second CD
and again, let's say the principal invested in each is $X.
![\bf a-b=3\qquad \implies \qquad \boxed{b}=3+a~\hfill \begin{cases} \left( \frac{a}{100} \right)X=240\\\\ \left( \frac{b}{100} \right)X=360 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \left( \cfrac{a}{100} \right)X=240\implies X=\cfrac{240}{~~\frac{a}{100}~~}\implies X=\cfrac{24000}{a} \\\\\\ \left( \cfrac{b}{100} \right)X=360\implies X=\cfrac{360}{~~\frac{b}{100}~~}\implies X=\cfrac{36000}{b} \\\\[-0.35em] ~\dotfill\\\\](https://tex.z-dn.net/?f=%5Cbf%20a-b%3D3%5Cqquad%20%5Cimplies%20%5Cqquad%20%5Cboxed%7Bb%7D%3D3%2Ba~%5Chfill%20%5Cbegin%7Bcases%7D%20%5Cleft%28%20%5Cfrac%7Ba%7D%7B100%7D%20%5Cright%29X%3D240%5C%5C%5C%5C%20%5Cleft%28%20%5Cfrac%7Bb%7D%7B100%7D%20%5Cright%29X%3D360%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cleft%28%20%5Ccfrac%7Ba%7D%7B100%7D%20%5Cright%29X%3D240%5Cimplies%20X%3D%5Ccfrac%7B240%7D%7B~~%5Cfrac%7Ba%7D%7B100%7D~~%7D%5Cimplies%20X%3D%5Ccfrac%7B24000%7D%7Ba%7D%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%20%5Ccfrac%7Bb%7D%7B100%7D%20%5Cright%29X%3D360%5Cimplies%20X%3D%5Ccfrac%7B360%7D%7B~~%5Cfrac%7Bb%7D%7B100%7D~~%7D%5Cimplies%20X%3D%5Ccfrac%7B36000%7D%7Bb%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C)


Answer:
7 mins
Step-by-step explanation:
Current speed of Joes Car = 65.5 mph
We have to find the time interval for which the car exceeded the speed limit of 55 mph.
While, we are given that the speed of the car was constantly increasing, hence the speed over all increased from the limit of 55 mph = 65.50-55.00 = 10.50 mph
We are also given that Joes car was increasing speed at a constant rate of 1.50 mph for every passing minute. Hence
1.50 mph is increased in 1 minute
1 mph will be increase in
minutes
Hence
10.50 mph will be increased in
minutes


Hence joes car was exceeding the limit of 55 mph for 7 minutes.
Let x be the smallest number.
x+(x+1)+(x+2)=384
3x+3=384
3x=381
x=127
The three consecutive numbers are 127, 128 and 129.
Hope my answer helped u :)
Answer:
t = -5
Step-by-step explanation:
Solve for t:
5 (t - 3) - 2 t = -30
Hint: | Distribute 5 over t - 3.
5 (t - 3) = 5 t - 15:
5 t - 15 - 2 t = -30
Hint: | Group like terms in 5 t - 2 t - 15.
Grouping like terms, 5 t - 2 t - 15 = (5 t - 2 t) - 15:
(5 t - 2 t) - 15 = -30
Hint: | Combine like terms in 5 t - 2 t.
5 t - 2 t = 3 t:
3 t - 15 = -30
Hint: | Isolate terms with t to the left hand side.
Add 15 to both sides:
3 t + (15 - 15) = 15 - 30
Hint: | Look for the difference of two identical terms.
15 - 15 = 0:
3 t = 15 - 30
Hint: | Evaluate 15 - 30.
15 - 30 = -15:
3 t = -15
Hint: | Divide both sides by a constant to simplify the equation.
Divide both sides of 3 t = -15 by 3:
(3 t)/3 = (-15)/3
Hint: | Any nonzero number divided by itself is one.
3/3 = 1:
t = (-15)/3
Hint: | Reduce (-15)/3 to lowest terms. Start by finding the GCD of -15 and 3.
The gcd of -15 and 3 is 3, so (-15)/3 = (3 (-5))/(3×1) = 3/3×-5 = -5:
Answer: t = -5