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Illusion [34]
3 years ago
15

Y - 18 = 2/7 (x + 6)

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
7 0

Answer:

y=2/7x+138/7

Step-by-step explanation:

hope this helps!

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Correct Answers only!
Airida [17]

Answer:

$15866

Step-by-step explanation:

Using the given formula :

I = P × R × T

I = 99160 × 8/100 × 2

I = $15866

7 0
3 years ago
PAULA BROUGHT $39.63 WORTH OF GROCERIES SHE GAVE TELLER TWO TWENTY DOLLARS BILLS WHAT IS THE FEWEST COIN SHE CAN GET IN CHANGE
MakcuM [25]

Answer:

4 coins

Step-by-step explanation:

First, let's find out how much she gave the teller:

2($20)=$40

Now, let's find out how much change the teller owes her:

$40.00-$39.63=$0.37 or 37 cents

The largest coin (under 37 cents), is a quarter which is worth 25 cents...

37-25=12 cents

The largest coin (under 12 cents), is a dime which is worth 10 cents...

12-10=2

The largest coin (under 2 cents), is a penny which is worth 1 cent...

2-1=1

And one more penny...

1-1=0

Therefore the fewest amount of coins she can get in change are a quarter, a dime, and two pennies, which is 4 coins.

7 0
2 years ago
Read 2 more answers
How much money should be invested at 5% interest, compounded quarterly, so that 7 years later the investment will be worth $5000
vampirchik [111]
\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$5000\\
P=\textit{original amount deposited}\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, four times}
\end{array}\to &4\\
t=years\to &7
\end{cases}
\\\\\\
5000=P\left(1+\frac{0.05}{4}\right)^{4\cdot 7}\implies 5000=P(1.0125)^{28}\\\\\\ \cfrac{5000}{(1.0125)^{28}}=P
8 0
3 years ago
I need help asap! Y over x minus 3 equals 6. I need help solving for Y.
monitta
That mean the awner to question is 3 y is 3 just substract 6 minus 3
6 0
3 years ago
Read 2 more answers
PLEASE HELP ME 79 POINTS!!!!!!!!!!!!!!!!!!
masya89 [10]

Answer:

Two squares with the same side lengths are always congruent.

Two quadrilaterals with the same side lengths are always congruent.

Two rectangles with the same side lengths are always congruent.

Step-by-step explanation:

7 0
2 years ago
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