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PtichkaEL [24]
3 years ago
12

NEED ANSWER ASAP!!!WILL VOTE BRAINLIEST ANSWER

Mathematics
1 answer:
Nikitich [7]3 years ago
5 0
I think
113.04 will be the answer.

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*cassually dances to “Savage Love” by Jason Durulo cause I’m ALMOST DONE!*
Montano1993 [528]

Answer:

savage love

Step-by-step explanation:

7 0
3 years ago
Which statement is true about f(x)-2/3lx+4l-6?
slamgirl [31]
1) Taking in account that the function is f(x)= -(2/3) |x+4|-6, I enclose a file with the graph.

That helps you to conclude:

a) The graph of f(x) has a vertex on (-4, -6)

b) When you multiply a function times 2/3 it is vertically compressed which is equivalent to horizontally streched.

c) The graph of f(x) opens downward

d) The domain of f(x) is all the real values (the absolute function accepts any value of x either positive or negative)

Answer: the graph of f(x) is horizontally stretched.

7 0
3 years ago
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How would I find the difference of 7/8 - 1/3 ?
Lemur [1.5K]

Answer:

\frac{13}{24}

Step-by-step explanation:

We have two fractions and are being asked to subtract them.

7/8 and 1/3 don't have the same denominator, so to do as so, we need to find the LCM of both denominators, multiply them to that, and do the same for the numerators.

So what is the LCM (Least Common Multiple) of 8 and 3?

24.

Therefore :

\frac{7 * 3}{8 * 3} - \frac{1*8}{3*8}

\frac{21}{24} - \frac{8}{24}

\frac{13}{24}

5 0
3 years ago
Read 2 more answers
Add and Subtract Rational Expressions with a Common Denominator
Sladkaya [172]

Answer:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}

Operating on the numerators:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}

Removing parentheses:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}

Simplifying:

\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}

The expression cannot be further simplified.

7 0
3 years ago
Help!!! Two customers took out car loans from a bank.
Nikitich [7]

Answer:

  • Susan paid $2220 more

Step-by-step explanation:

<u>Use the interest formula:</u>

  • I = Prt, where P - amount of loan, r- interest rate, t- time in years

<u>Robert:</u>

  • I = 30000*(4.9/100)*4 = 5880

<u>Susan:</u>

  • I = 30000*(4.5/100)*6 = 8100

<u>Difference in amounts of interest:</u>

  • 8100 - 5880 = 2220

Susan paid $2220 more

8 0
3 years ago
Read 2 more answers
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