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qwelly [4]
2 years ago
13

Solve -3 < 2x + 3 < 5

Mathematics
1 answer:
lana [24]2 years ago
7 0

The solution of the given Inequality i.e. -3<2x+3≤5 is x > -3 and x ≤1 (option B).

Inequality:

When the symbols ">", "<", "≤", or "≥" are used to connect two real numbers or algebraic expressions, that relationship is known as an Inequality.

Here the given Inequality is -3<2x+3≤5

⇒ -3<2x+3≤5

⇒-3-3 < 2x+3-3 ≤ 5-3 { add -3 on both the sides }

⇒-6 < 2x ≤ 2

⇒-3 < x ≤ 1

⇒-3 < x and  x ≤ 1

⇒ x > -3 and x ≤1

Therefore,The solution of the given Inequality i.e. -3<2x+3≤5 is x > -3 and x ≤1 (option B).

Learn more about Inequality here:

brainly.com/question/20383699

#SPJ4

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This is quite easy, but anyway:

In order, they are

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7 - 3 = 4

4 would be the range
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Steve has $20.00 to spend. He decided to buy a hat for $12.75 and spend the rest on candy bars. If each candy bar costs $1.25, h
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Answer:

5

Step-by-step explanation:

To find how many candy bars Steve can buy, we need to find the total amount of money Steve has for buying candy.

Amount of money for candy bars = $20.00 - $12.75

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Two water pumps are filling a pool. One of the pumps is high power and can fill the pool 5 hours before the other can do. Howeve
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L = hours for the slower pump to fill the pool

L - 5 = hours for the high power pump, since it can do it in 5 hours less.

since the slower pump takes L hours, the faster pump takes then L - 5 hours.

we know both pumps together take 3 hours to do <u>half of the pool</u>, so that means that to fill up <u>the whole pool it takes them 6 hours</u>.

since the slower pump can do it alone in L hours, in 1 hour it has done 1/L of the whole thing.

likewise, since the faster pump can do it in L-5 hours, in 1 hour alone it has done 1/(L-5) of the whole job.


\bf \stackrel{\textit{slower pump's rate}}{\cfrac{1}{L}}~~~~+\stackrel{\textit{faster pump's rate}}{\cfrac{1}{L-5}}~~~~=~~~~\stackrel{\textit{total done in 1 hour}}{\cfrac{1}{6}} \\\\\\ \textit{let's multiply both sides by }\stackrel{LCD}{6(L)(L-5)}\textit{ to do away with the denominators} \\\\\\ 6(L)(L-5)\left( \cfrac{1}{L}+\cfrac{1}{L-5} \right)=6(L)(L-5)\left( \cfrac{1}{6} \right)


\bf 6(L-5)+6L=(L)(L-5)\implies 6L-30+6L \\\\\\ 12L-30=L^2-5L\implies 0=L^2-17L+30 \\\\\\ 0=(L-15)(L-2)\implies L= \begin{cases} \boxed{15}\\ 2 \end{cases}


now, we can't use L = 2, because, the whole job by both is done in 6 hours, there's no way the slower pump can do it in less than that, so L = 15.

now if L = 15, then L - 5 = 10.

5 0
4 years ago
a group of four students is performing an experiment with salt each student must a 3/8 teaspoon of salt to a solution the group
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Answer:

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The next step is to calculate the area that the entire parade is occupying. This can be done by multiplying the length of the parade by the breadth. In this case, it will be 21120ft X 5 ft = 105600 square feet.

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