2 years ago

PLZZ HELP WILL MARK BRAINLIST!!!!

Mathematics

1 answer:

DedPeter [7]2 years ago

4 0

X intercept- (0,250)
Y intercept- (0,100)
If you want me to explain I can!

You might be interested in

If XY= 30, and XZ= 100, then YZ= ____.

marishachu [46]

Answer:

yz= 100-30

yz= 70.

hope it helps.

3 0

2 years ago

Tony wants to save $10,000 in 6 years. Assuming a 4% interest rate, what is the minimum he must save each month to reach his goa

arlik [135]

We have been given that Tony wants to save $10000 in 6 years.

That means future value S = $10000

Time t= 6 years

Interest rate = 4% yearly = 0.04 yearly

n=12 months per year

Now we have to find monthly payment to recieve $10000 in 6 years. so we need to apply monthly payment formula which is

$S=R(\frac{(1+\frac{r}{n})^{nt}-1}{\frac{r}{n}})$

$10000=R(\frac{(1+\frac{0.04}{12})^{12*6}-1}{\frac{0.04}{12}})$

$10000=R(\frac{(1+0.00333333333333)^{72}-1}{0.00333333333333})$

$10000=R(\frac{(1.00333333333333)^{72}-1}{0.00333333333333})$

$10000=R(\frac{1.27074187908-1}{0.00333333333333})$

$10000=R(\frac{0.27074187908}{0.00333333333333})$

$10000=R(81.2225637241)$

$\frac{10000}{81.2225637241}=R$

$123.11849739=R$

which is approx $123.

Hence final answer is C) $123.

4 0

3 years ago

sarah is taking out 14000 four year new car loan with an apr of 6.75% what is the monthly payment for this loan

jeka57 [31]

Answer:

Monthly payment = $18325.6

Step-by-step explanation:

From the question,

P = 14000

t = 4 years

R = 6.75%

To calculate the monthly payment for this loan, let's use the below formula

Amount = P( 1 + 6.75%/12)^4×12

Amount = 14000(1.005625)^48

Amount = 18325.633 dollars

Monthly payment = $18325.6

6 0

3 years ago

A student bought tickets online the total cost c in dollar of t, tickets can be found using the fuction below c=24.50t+9.50

dalvyx [7]

What is the total cost or it is unsolvable?

8 0

3 years ago

Can someone pls help ASAP :) I’ll give brainiest

Anna71 [15]

Answer:

D. (2x)(3x) + 5(3x - 2)

<em>good luck, i hope this helps :) </em>

<em />

5 0

2 years ago