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alexdok [17]
3 years ago
9

Help please I'm really not getting this for some reason

Mathematics
2 answers:
BartSMP [9]3 years ago
8 0
The first given equation is:
x - 7y = -20
We can rewrite this equation as follows:
x = 7y - 20 ...........> equation I

The second given equation is:
-3x + 9y = 36 ..........> equation II

Substitute with equation I in equation II to get the value of the y as follows:
-3x + 9y = 36
-3(7y-20) + 9y = 36
-21y + 60 + 9y = 36
-12y = 36-60 = -24
y = (-24) / (-12) = 2

Now use the value of the y and substitute in equation I to get the value of the x as follows:
x = 7y - 20
x = 7(2) - 20 = 14 - 20 = -6

Based on the above calculations:
x = -6
y = 2
Bingel [31]3 years ago
8 0
Original system:
x - 7y = -20
-3x + 9y = 36

The two methods of solving systems of equations are elimination and substitution. It looks like substitution will be easier here, so I'm going to use that method. I'll take the first term and isolate x, like so:

x - 7y = -20         (original equation)
x - 7y + 20 = 0    (subtract 20 from both sides)
-7y + 20 = -x       (subtract x from both sides)
7y - 20 = x           (divide both sides by -1)

Now what we can do it plug this expression into the second equation for x and solve for y:

-3(7y - 20) + 9y = 36    (substitute the expression for x)
-21y + 60 + 9y = 36      (distribute the -3)
-21y + 9y = -24             (subtract 60 from both sides)
-12y = -24                     (combine like terms)
y = 2                              (divide both sides by -12)

Now that we have y, we can substitute it back into the first equation to find x:

x - 7(2) = -20    (substitute your value for y into the equation)
x - 14 = -20       (multiply)
x = -6                (add 14 to both sides)

Finally, we must check our work by plugging our values back into both of the original equations:

-6 - 7(2) = -20     -->    -6 - 14 = -20    <--This is correct
-3(-6) + 9(2) = 36    -->    18 + 18 = 36    <--This is correct

If you have any question about this process, let me know.
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Answer:

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3 years ago
Which statement is true about the domain of y = 3(2^-x)? Explain.
SVEN [57.7K]
We want to determine the domain of {y=3 \cdot 2^{-x}=3 \cdot ({2^{-1}})^x=3 \cdot ({ \frac{1}{2}})^x

any function of the form y=f(x)=a \cdot b^x is called an "exponential function",
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That is for any x, the expression <span>3(2^-x) "makes sense".



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8 0
3 years ago
Please answer this. thank you
Rama09 [41]

Answer:  (-\infty, -4]

Curved parenthesis at negative infinity

Square bracket at -4

====================================================

Work Shown:

5(x+4) \le 0 \\\\x+4 \le \frac{0}{5} \\\\x+4 \le 0 \\\\x \le 0-4 \\\\x \le -4 \\\\

The last inequality shown above is the same as saying -\infty < x \le -4

Converting this to interval notation leads to the final answer of (-\infty , -4]

Note the use of a square bracket at -4 to include this endpoint. We can never include either infinity, so we always use a parenthesis for either infinity.

5 0
3 years ago
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Vanyuwa [196]

Answer:

11/24

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61 / 24 +  x = 3

x = 72/24 - 61/24

x = 11 / 24

Tim has to run 11/24 of the race

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3 years ago
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elena-s [515]
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That is the basic definition of what supplementary angles are and can be found in any textbook.
4 0
3 years ago
Read 2 more answers
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