8. The equation at the right has been incorrectly solved.

Mathematics

2 answers:

ozzi2 years ago

8 0

The equation at the right has been incorrectly solved, the answer to this is b = -7

Lilit [14]2 years ago

6 0

When subracting the negative 24 from the left and right side, the left will cancel and the right will equal -28 instead of positive.

The final answer would just be positive 7 instead of negative.

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A shoebox is covered with packing paper before being shipped. The paper cost $0.02 per square inch. How much will it cost to cov

Yuliya22 [10]

It will cost about $0.72 for three pair of shoes assuming shoe box is rectangular shape having 12 square inches of packing paper.

Step-by-step explanation:

Assuming a shoe box rectangle in shape and the same size is required.
Length of rectangle box assuming to be 4 inches.
Width of rectangle box assuming to be 3 inches.
as we know area of rectangle is Length * Breath(Width) .
12 square inches is the total packing paper required for each box.
To cover cost of three pairs of shoes $0.02 per square inch.
Each box would cost about 12 square inches * 0.02 pr square inches.
0.24 square inches of packing paper will be required.
let three pairs of shoes be x
Total pairs of shoes would be= (12 * 0.02)*x
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3 0

3 years ago

If 6x = 1/126 find the value of x.

Anastasy [175]

Answer:

x = $\frac{1}{756}$

Step-by-step explanation:

Given

6x = $\frac{1}{126}$ ( divide both sides by 6 )

x = $\frac{\frac{1}{126} }{6}$ = $\frac{1}{126}$ × $\frac{1}{6}$ = $\frac{1}{126(6)}$ = $\frac{1}{756}$

6 0

3 years ago

Read 2 more answers

H(x)=1/8x^3-x^2 Over which interval does h have a positive average rate of change?

lara31 [8.8K]

Answer:

$(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}$

Step-by-step explanation:

$h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x$

For positive rate of change $h'(x)>0$

$\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\$