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4 years ago

Help me pleeeeeeeeeaze

Mathematics

1 answer:

kobusy [5.1K]4 years ago

5 0

$a(x+3) < 5x+15-x\\\\ax+3a < 5x+15-x\\\\ax+3a < 4x+15\ \ \ |-3a\\\\ax < 4x+15-3a\ \ \ \ |-4x\\\\ax-4x < 15-3a\\\\(a-4)x < 15-3a\\\\\text{The solution is all real numbers if}\ a-4=0\ \ \ |+4\\\\\boxed{a=4}\\\\\text{Check:}\\\\4(x+3) < 5x+15-x\\4x+12 < 4x+15\ \ \ \ |-4x\\12 < 15\ TRUE\to x\in\mathbb{R}$

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Solve each following addition and subtraction problems. 82+25+92=

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Answer:

199

Step-by-step explanation:

82+25=107

107+92= 199

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3 years ago

Is 3/5 bigger than 65%

Effectus [21]

If we think 1(5/5) is 100%
3/5=100×3/5=60%
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The box plot summarizes the number of people enrolling in fitness classes each month during one year at the city gym. The mean o

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Answer:

first blank: mean

second blank: greater than

third blank: increased

fourth blank: greater than

Step-by-step explanation:

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3 years ago

Read 2 more answers

A 10-foot ladder leans against a wall so that it is 6 feet high at the top. The ladder is moved so that the base of the ladder t

konstantin123 [22]

Answer:

The top of the ladder is now at 10 ft.

Step-by-step explanation:

At the start, we have a height H=6, a length L=10 and a base B, that has to be calculated by the Pythagorean theorem:

$B^2=L^2-H^2=10^2-6^2=100-36=64\\\\B=\sqrt{64}=8$

The base is moved twice the distance the height moves up.

We called this distance x, so we have:

$L^2=(H+x)^2+(B-2x)^2=H^2+2Hx+x^2+B^2-4Bx+4x^2\\\\L^2=(H^2+B^2)+5x^2+(2H-4B)x\\\\L^2=L^2+5x^2+(2H-4B)x\\\\0=5x^2+(2H-4B)x\\\\5x+(2H-4B)=0\\\\x=\dfrac{4B-2H}{5}=\dfrac{4*8-2*6}{5}=\dfrac{32-12}{5}=\dfrac{20}{5}=4$

The new height (H+x) is

$H'=H+x=6+4=10$

The base travels 2x=8, so the new base B' is 0.

This means that the ladder is all against the wall (L=H').

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3 years ago

HELP!!! i need it asap, thank you

Step2247 [10]

Answer: H

Step by Step

7 0

4 years ago