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3 years ago

5

Which expression correctly represents three less than the product of a number and two increased by five?

1 answer:

daser333 [38]3 years ago

5 0

Answer:

The answer is 3-n2-5

<3

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How can 52/9 be expressed as a decimal? <br><br> <br> <br> <br>

slava [35]

The answer would be about 5.8

4 0

3 years ago

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What is 1.08 to the power of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D" id="TexFormula1" title="\frac{1}{5}" alt="

OlgaM077 [116]

Answer:

$about \: 1.01551$

Step-by-step explanation:

No it isn't, because:

${1.08}^{ \frac{1}{5} } = { \frac{108}{100} }^{ \frac{1}{5} } = { \frac{27}{25} }^{ \frac{1}{5} } = \sqrt[5]{ \frac{27}{25} } = 1.01551$

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3 years ago

Find the lcd. Add or subtract the fractions. Simplify if needed. (- 3/16) + (- 5/2)

Solnce55 [7]

Answer:

$\boxed{-\frac{43}{16}}$

.

Step-by-step explanation:

$-\frac{3}{16} + (-\frac{5}{2} )$

$= -\frac{3}{16} - \frac{5}{2}$

$= \frac{-3 - 5(8)}{16}$

$= \frac{-3 - 40}{16}$

$= -\frac{43}{16}$

8 0

3 years ago

2. Determine the sum of the first 400 ODD numbers.<br><br>

il63 [147K]

Odd numbers take the form $2n-1$ , where $n\ge1$ is an integer. When $n=400$ , the last odd number would be 799. So we're adding

$S=1+3+5+\cdots+795+797+799$

By reversing the order of terms, we have

$S=799+797+795+\cdots+5+3+1$

and we can pair up terms in both sums at the same position to write

$2S=(1+799)+(3+797)+(5+795)+\cdots(795+5)+(797+3)+(799+1)$

so that we are basically adding 400 copies of 800, and from there we can find the value of the sum right away:

$2S=400\cdot800\implies S=160,000$

###

We could also make use of the formulas,

$\displaystyle\sum_{i=1}^n1=n$

$\displaystyle\sum_{i=1}^ni=\dfrac{n(n+1)}2$

We have

$S=\displaystyle\sum_{i=1}^{400}(2i-1)=2\sum_{i=1}^{400}i-\sum_{i=1}^{400}1=400(400+1)-400=400^2=160,000$

3 0

3 years ago

Help i need tis anseer quick

lesya692 [45]

Answer:

9

Step-by-step explanation:

can fit in 9 times

4 0

3 years ago

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