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Maurinko [17]
3 years ago
8

Please help question in the picture

Mathematics
1 answer:
zubka84 [21]3 years ago
7 0
Its 56 I DID this you could only multiple the straight sides
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Solve the system using elimination.<br> 2x + 3y = -2<br> 3х – 6у = 18
Mashutka [201]

Answer:

x = 2 and y = -2

Step-by-step explanation:

We must first eliminate the y-variable:

2x + 3y = -2

3x - 6y = 18

4x + 6y = -4

3x - 6y = 18

_________

7x = 14

x = 2

Since we know x=2, we plug that back into one of the original equations to find y:

2x + 3y = -2

2(2) + 3y = -2

4 + 3y = -2

3y = -6

y = -2

Therefore, x = 2 and y = -2

7 0
3 years ago
Read 2 more answers
What is the length of the hypotenuse,x,if (20,21,x) is a pythogorean triple?
pickupchik [31]

check the picture below.

3 0
3 years ago
James has a balance of $1700 on a credit card with an APR of 18.7%, compounded monthly. About how much will he save in interest
miss Akunina [59]

Answer:

  $121.51

Step-by-step explanation:

At 18.7%, the monthly multiplier is 1+.187/12, so for the year, James' balance is multiplied by (1+.187/12)^12 ≈ 1.2038899

At 12.5%, the monthly multiplier is 1+.125/12, so for the year, James' balance is multiplied by (1+.125/12)^12 ≈ 1.1324161

The difference in these multipliers is ...

  1.2038899 -1.13241605 = 0.0714739

so James' savings is ...

  $1700 × 0.0714739 ≈ $121.51

5 0
3 years ago
What is −5/8÷9 1/0 ? <br> −25/36 a<br> −91/6 b<br> −35/80 c<br> −1/36 d
Alla [95]

Answer:

23

Step-by-step explanation:

3 0
3 years ago
(x+7)/(x^2-49) find the domain. show work
harina [27]
\large\begin{array}{l} \textsf{Find the domain of}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-49}}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-7^2}}\\\\\\ \textsf{Factor out the denominator using special products:}\\\\ \textsf{(a difference of squares)}\\\\ \mathsf{f(x)=\dfrac{x+7}{(x+7)(x-7)}} \end{array}


\large\begin{array}{l} \textsf{Restrictions for the domain:}\\\\ \bullet~~\textsf{Denominators must not be zero:}\\\\ \mathsf{(x+7)(x-7)\ne 0}\\\\ \begin{array}{rcl} \mathsf{x+7\ne 0}&~\textsf{ and }~&\mathsf{x-7\ne 0}\\\\ \mathsf{x\ne -7}&~\textsf{ and }~&\mathsf{x\ne 7} \end{array} \end{array}


\large\begin{array}{l} \textsf{Therefore, the domain of f is}\\\\ \mathsf{D_f=\{x\in\mathbb{R}:~~x\ne -7~~and~~x\ne 7\}}\\\\\\ \textsf{or using a more compact form}\\\\ \mathsf{D_f=\mathbb{R}\setminus\{-7,\,7\}}\\\\\\ \textsf{or using the interval notation}\\\\ \mathsf{D_f=\left]-\infty,\,-7\right[\,\cup\,\left]7,\,+\infty\right[.} \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2155752


\large\textsf{I hope it helps. :-)}



Tags: <em>function domain real rational factorizing special product interval</em>

</span>
7 0
3 years ago
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