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seropon [69]
3 years ago
11

Please help I need this done by 5:00​

Mathematics
1 answer:
Aliun [14]3 years ago
6 0
<h2>PLEASE DOUBLE CHECK I AM SO OUT OF PRACTICE WITH THESE OMG</h2>

A. a = 25º

B. 41º

C. 41º

uh werk:

180º = 2a - 9º + 5a + 14º

180º = 7a + 5º

175º = 7a

a = 25º

180 - (5a+14) = 180 - 139 = 41

or 2a-9 = 41

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Samuel and Christopher go to the movie theater and purchase refreshments for their friends. Samuel spends a total of $34.00 on 6
Llana [10]

Answer:

The system of equations is :

6 p + 4 d = 34

12 p + 7 d = 66.25

The price of a drink is $1.75

Step-by-step explanation:

Assume that the price of one bag of popcorn is $p and the price of a drink is $d

∵ Samuel spends a total of $34 on 6 bags of popcorn and

   4 drinks

- Multiply p by 6 and d by 4, then add the products and equate

    the sum by 34

∴ 6 p + 4 d = 34 ⇒ (1)

∵ Christopher spends a total of $66.25 on 12 bags of popcorn

    and 7 drinks

- Multiply p by 12 and d by 7, then add the products and equate

    the sum by 66.25

∴ 12 p + 7 d = 66.25 ⇒ (2)

The system of equations is :

6 p + 4 d = 34

12 p + 7 d = 66.25

Now let us solve it

Multiply equation (1) by -2 to make the coefficients of p in the two equations equal in values and opposite in signs to eliminate it

∵ -12 p - 8 d = - 68 ⇒ (3)

- Add equations (2) and (3)

∴ - d = - 1.75

- Divide both sides by -1

∴ d = 1.75

The price of a drink is $1.75

7 0
3 years ago
8^3x=16^2x<br> Can you please solve this it is 8 to the power of 3x and 16 to the power of 2x
GarryVolchara [31]

Answer:

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3 0
2 years ago
What's the anewer to this problem..?
valentina_108 [34]
21x(9)=9x21
21y= 9x21
Y=9x21/21
Y=9x1
Y=9
4 0
3 years ago
Read 2 more answers
At the beginning of year 1, Sam invests $700 at an annual compound interest
Aleksandr-060686 [28]

at the beginning of year 4, only 3 years have elapsed, the 4th year hasn't started yet, since it's at the beginning, so at the beginning of year 4 we can say only 4-1 years have elapsed.

~~~~~~ \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}\dotfill &\$700\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=\textit{elapsed years}\dotfill &3 \end{cases}

A=700\left( 1 + \frac{0.05}{1} \right)^{1\cdot 3}\implies A = 700(1+0.05)^3\implies A(4)=700(1+0.05)^{4-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A(n)=700(1+0.05)^{n-1}~\hfill

6 0
2 years ago
4 – 2(x + 7) = 3(x + 5)
galina1969 [7]

-5 is your answer

steps: 4-5 then you distribute and combine and you should get -5

4 0
4 years ago
Read 2 more answers
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