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3 years ago

Marking brainliest!!!

Mathematics

2 answers:

Furkat [3]3 years ago

7 0

11 A
12 C
13 1
14 D
15 B
16 2
17 C
18 9

arlik [135]3 years ago

5 0

11 OPTION IS A

12 OPRION IS C

13 ANSWER IS 1

14 OPTION IS D

15 OPTION IS B

16 ANSWER IS 2

17 OPTION IS C

18 ANSWER IS 9

pls mark me as the brainliset hope it helps you

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Find the inverse of the function y= -3/ x+ 4

Tomtit [17]

Y=4-3/x
y-4=-3/x
(y-4)x=-3
x(y-4)=-3
x=-3/y-4

there's you answer

7 0

3 years ago

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3. When 25 is added to 7 times a number, the result is 81. Find the number.

Dominik [7]

Answer:

3. 2.5

4. $5 per candy bar

5. You mowed 6 lawns

6. 15.66 is the number

7. 13 weeks

8. 135 students came

9. 12

Step-by-step explanation:

3. 25+7= 32.

81/32= 2.5

4. 23-3= 20

20/4= 5

5. 155-35=120

120/20=6

6. 10-4 = 6

94/6=15.66

(not 100% sure about this one)

7. 200-161=39

39/3=13

8. 975-300=675

675/5=135

9. 36-12=24

36/3= 12

(not 100% sure about this one)

Bonus: Sorry I don't know

I hope I can help! Also next time do each question one by one so other people can answer and not answer all of them if they don't know one.

4 0

3 years ago

Find the value of $ 15,000 at the end of one year if it is invested in an account that has an interest rate of 4.95 % and is com

lora16 [44]

A)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{twelve months, thus}
\end{array}\to &12\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{12}\right)^{12\cdot 1}$

b)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{365 days, thus}
\end{array}\to &365\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{365}\right)^{365\cdot 1}$

c)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{four quarters, thus}
\end{array}\to &4\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{4}\right)^{4\cdot 1}$

7 0

3 years ago

What is the m∠ACB? 10° 50° 90° 180°

Musya8 [376]

Answer:

Step-by-step explanation:

ßimple answer no needs to panic

7 0

1 year ago

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Which of the following are factors of P ( x ) = 2 x 3 + 3 x 2 − 8 x − 12 ?

alukav5142 [94]

Answer:

108

Step-by-step explanation:

12-8(-12)

12-(-96)

12+96

108

6 0

2 years ago

Read 2 more answers