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Strike441 [17]
3 years ago
5

3x~6y=~3 2x+4y=30 Elimination using multiplication

Mathematics
1 answer:
Nat2105 [25]3 years ago
7 0
\left \{ {{3x-6y=-3\ \ |*2} \atop {2x+4y=30\ \ | *(-3)}} \right. \\\\ \left \{ {{6x-12y=-6} \atop {-6x-12y=-90}} \right. \\+----\\elimination\\\\
0 \neq -69\\\\There\ are\ no\ solutions
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A collection of quarters and nickels is worth $14.80. There are 88 coins in all.
ycow [4]
So the equaiton is

number of nickes=n
number of quarters=q

if you have 1 quarter then you have 25 cents so we will represent like this
25q

and nickles is 5n
14.50=1450 cents

so
25q+5n=1480
q+n=88
subtract q from both sides
n=88-q
subsitute into first equation
25q+5(88-q)=1480
25q+440-5q=1480
add like terms
20q+440=1480
subtract 440 from both sides
20q=1040
divide both sdies by 20
q=52
there were 52 quarters
25(52)+5n=1480
1300+5n=1480
subtract 1300 from both sides
5n=180
divide both sides by 5
n=36
there were 36 nickels


nickles=36
quarters=52
8 0
4 years ago
Question in above picture<br> explanation please not jusr answer
Lena [83]

Answer:

x+16

Step-by-step explanation:

For this one the negative signs become positive because, if you multiply a negative number*a negative number it becomes positive. That is a rule.

-(-x-16)\\-1(-1x)-1(-6)\\x+16

7 0
3 years ago
Read 2 more answers
In the following calculation, which part should be calculated first? 3 + (2 * 4) - 5 / 2
meriva

Always solve the parenthese first

8 0
3 years ago
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How would I Factor x²+2x-7
kolezko [41]

Answer:

It cannot be factored

Step-by-step explanation:

There isn’t any number that multiplies to -7 and add to 2

3 0
3 years ago
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What is the slope of a line perpendicular to the line whose equation is 2x+3y=21. Fully simplify your answer.
erastova [34]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

2x+3y=21\implies 3y=-2x+21\implies y=\cfrac{-2x+21}{3} \\\\\\ y=\cfrac{-2x}{3}+\cfrac{21}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+7\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

therefore then

\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-2}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{-2}\implies \cfrac{3}{2}}}

3 0
2 years ago
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