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3 years ago

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Mathematics

2 answers:

Daniel [21]3 years ago

5 0

Answer: the first one

Step-by-step explanation:

Kaylis [27]3 years ago

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Answer:

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Step-by-step explanation:

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What is the answer to this problem 3/2 x=−15

fomenos

Answer:

-15×2=-30

30÷3=10

not sure

4 0

3 years ago

Abel was wrapping a gift for an upcoming party. He began with 20 1/2 feet of paper. He used 5 5/12 feet on his first gift and 3

FrozenT [24]

Hey there!

$20 \dfrac{1}{2} - 5 \dfrac{5}{12} - 3 \dfrac{1}{3} + 2 \times 10 \dfrac{1}{2} \\ = \dfrac{41}{2} - \dfrac{65}{12} - \dfrac{10}{3} + 21 \\ = \dfrac{246}{12} - \dfrac{65}{12} - \dfrac{40}{12} + 21 \\ = \dfrac{141}{12} + 21 \\ = 32 \dfrac{3}{4}$

Answer : 32 3/4 feet or in decimal form, 32.75 feet.

Hope this helps. - M

5 0

3 years ago

Read 2 more answers

A box measures 15 inches by 8 inches by 5 inches. What is the length of its longest diagonal?

Bond [772]

17 inches

15^2 + 8^2 = 289
square root of 289 = 17 inches

5 0

4 years ago

A=.5h(B+b); A=15, B=2, b=3 <br> it's asking for the unknown variable

PtichkaEL [24]

Answer:

Step-by-step explanation:

That is the area of a trapezoid

A=h(b1+b2)/2

We are given everything but the height so

h=(2A)/(b1+b2), now using the values given to us we have

h=(2(15))/(2+3)

h=30/5

h=6

3 0

3 years ago

Combine into a single logarithm.<br><br> 3log(x+y)+2log(x-y)-log(x^2 +y^2)

seropon [69]

Answer:

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Step-by-step explanation:

Given the expression

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

solving to write into a single logarithm

$3log\left(x+y\right)+2log\left(x-y\right)-log\left(x^2\:+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$3\log _{10}\left(x+y\right)=\log _{10}\left(\left(x+y\right)^3\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\log _{10}\left(x-y\right)=\log _{10}\left(\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _{10}\left(\left(x+y\right)^3\right)+\log _{10}\left(\left(x-y\right)^2\right)=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)$

$=\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _{10}\left(\left(x+y\right)^3\left(x-y\right)^2\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

$=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

Thus,

$3\log _{10}\left(x+y\right)+2\log _{10}\left(x-y\right)-\log _{10}\left(x^2+y^2\right)=\log _{10}\left(\frac{\left(x+y\right)^3\left(x-y\right)^2}{x^2+y^2}\right)$

6 0

3 years ago