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Anuta_ua [19.1K]
3 years ago
12

Solve the inequality.

Mathematics
1 answer:
gulaghasi [49]3 years ago
6 0

\\ \sf\longmapsto q+12-2(q-22)>0

\\ \sf\longmapsto q+12-2q+44>0

\\ \sf\longmapsto q-2q+12+44>0

\\ \sf\longmapsto -q+56>0

\\ \sf\longmapsto -q>-56

\\ \sf\longmapsto q

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Will mark as brainliest plz helpp​
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Step-by-step explanation:

x - 4 = √3y

x - 4 <u>- √3y</u> = √3y <u>- √3y</u>

x - 4 - √3y = 0

x - √3y - 4 = 0

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umka21 [38]

c = cost per pound of chocolate chips

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\bf \stackrel{\textit{3 lbs of "c"}}{3c}+\stackrel{\textit{5 lbs of "w"}}{5w}~~=~~\stackrel{\textit{costs}}{15} \\\\\\ \stackrel{\textit{12 lbs of "c"}}{12c}+\stackrel{\textit{2 lbs of "w"}}{2w}~~=~~\stackrel{\textit{costs}}{33} \end{cases}\qquad \impliedby \textit{let's use elimination} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{llccccccl} 3c+5w=15&\times (-4)\implies &-12c&+&-20w&=&-60\\ 12c+2w=33&&12c&+&2w&=&33\\ \cline{3-7}\\ &&0&&-18w&=&-27 \end{array}

\bf -18w=-27\implies w=\cfrac{-27}{-18}\implies \blacktriangleright w=\cfrac{3}{2} \blacktriangleleft \\\\\\ \stackrel{\textit{substituting on the 1st equation}}{3c+5\left(\cfrac{3}{2} \right)=15}\implies 3c+\cfrac{15}{2}=15 \implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( 3c+\cfrac{15}{2} \right)=2(15)} \\\\\\ 6c+15=30\implies 6c=15\implies c=\cfrac{15}{6}\implies \blacktriangleright c=\cfrac{5}{2} \blacktriangleleft

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Answer:

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15 mins = 1/4 of full circle

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If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

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Hi I think your answer should be 6.28 yards
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