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3 years ago

3^4+4.5= Input whole number

Mathematics

2 answers:

Nina [5.8K]3 years ago

7 0

Step-by-step explanation:

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Rasek [7]3 years ago

3 0

Answer:

thats 86 because it's the nearest whole number

Step-by-step explanation:

(3 × 3 × 3 × 3) + 4.5

81 + 4.5

85.5

<h2><em><u>SO THAT MEANS 3^4+4.5 = 85.5 rounded to the nearest whole number is 86</u></em></h2>

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What is 2x divided by 2

boyakko [2]

<u>2x
</u> 2

The 2's will reduce down to 1's

<u>1x</u>
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7 0

3 years ago

Help please I can’t figure it put

leonid [27]

5y + 3 > -7 y + 13

Subtract 3 from both sides

5y > -7y + 13 - 3

Add 7y to both sides

5y + 7y > 13 - 3

12y > 10

Divide by 12, which is positive so the sign doesn't flip

y > 10/12 = 5/6

Answer: y > 5/6

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3 years ago

What is The gcf of 13 and 7

algol13

Answer:

Step-by-step explanation:

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3 years ago

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GaryK [48]

Answer:

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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