The equation of the central street PQ is -1.5x - 3.5y = -31.5 option (b) is correct.
<h3>What is a straight line?</h3>
A straight line is a combination of endless points joined on both sides of the point.
We have a straight line:
Convert it to the general form given below:
or
(Slope of AB line)
For the slope(m') of the PQ line:
( because AB and PQ are perpendicular to each other)
We know the:
Where (x', y') = (7, 6), we get:
(multiply by -1/2 on both sides)
Thus, the equation of the central street PQ is -1.5x - 3.5y = -31.5
Learn more about the straight line.
brainly.com/question/3493733
Answer:
85 miles per hour
Step-by-step explanation:
9514 1404 393
Answer:
- late only: 15
- extra-late only: 24
- one type: 43
- total trucks: 105
Step-by-step explanation:
It works well when making a Venn diagram to start in the middle (6 carried all three), then work out.
For example, if 10 carried early and extra-late, then only 10-6 = 4 of those trucks carried just early and extra-late.
Similarly, if 30 carried early and late, and 4 more carried only early and extra-late, then 38-30-4 = 4 carried only early. In the attached, the "only" numbers for a single type are circled, to differentiate them from the "total" numbers for that type.
__
a) 15 trucks carried only late
b) 24 trucks carried only extra late
c) 4+15+24 = 43 trucks carried only one type
d) 38+67+56 -30-28-10 +6 +6 = 105 trucks in all went out
the cheap answer is simply
(x-5)(x²+4x-2)
we can simply multiply the terms on one by the terms of the other and then add like-terms and simplify.
![\bf (x-5)(x^2+4x-2)\implies \begin{array}{cllll} x^2+4x-2\\ \times x\\ \cline{1-1}\\ x^3+4x^2-2x \end{array}+ \begin{array}{cllll} x^2+4x-2\\ \times -5\\ \cline{1-1}\\ -5x^2-20x+10 \end{array} \\\\\\ x^3+4x^2-2x-5x^2-20x+10\implies x^3+4x^2-5x^2-2x-20x+10 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill x^3-x^2-22x+10~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%28x-5%29%28x%5E2%2B4x-2%29%5Cimplies%20%5Cbegin%7Barray%7D%7Bcllll%7D%20x%5E2%2B4x-2%5C%5C%20%5Ctimes%20x%5C%5C%20%5Ccline%7B1-1%7D%5C%5C%20x%5E3%2B4x%5E2-2x%20%5Cend%7Barray%7D%2B%20%5Cbegin%7Barray%7D%7Bcllll%7D%20x%5E2%2B4x-2%5C%5C%20%5Ctimes%20-5%5C%5C%20%5Ccline%7B1-1%7D%5C%5C%20-5x%5E2-20x%2B10%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5C%5C%20x%5E3%2B4x%5E2-2x-5x%5E2-20x%2B10%5Cimplies%20x%5E3%2B4x%5E2-5x%5E2-2x-20x%2B10%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20x%5E3-x%5E2-22x%2B10~%5Chfill)
Answer:
Ans. you should deposit each month to end up with $38,000 the amount of $533.33 every month for 5 years at a APR of 6.5%
Step-by-step explanation:
Hi, first we have to convert all the data to monthly basis, that is, 5 years (5*12=60 months) and the rate of 6.5% APR offered by the bank (Monthly rate = 0.065/12=0,005666667 or 0.5667% monthly)
With that in mind, we need to solve for "A" the following equation.
Where:
FV = Future value of the car
r = rate of return offered by the bank
n = number of periods that you are going to make the monthly deposit
That is:
Best of luck.
