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3 years ago

Decide whether the following situation is looking for the percent, the part, or the whole:

Mathematics

1 answer:

shtirl [24]3 years ago

5 0

The 14 compact cars sold by a dealer on Monday was 56% of the cars sold that day.

Situation is looking for a percent of compact car that is 56 or part that is 14 compact car out of whole.

<u>Solution: </u>

Given that

14 compact cars sold by a dealer on Monday was 56% of the cars sold that day.

Need to decide whether the following situation is looking for the percent, the part, or the whole.

From given information ,

The part that is 14 compact cars out of all cars are sold on the day.

Also 14 compact car are 56% of the car sold on that day.

So clearly situation is looking for a percent of compact car that is 56 or part that is 14 compact car out of whole.

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Pls help me this and explain for me pls

Phoenix [80]

Answer:

m<2 is 45° because there is proof in the problem.

m<1 is 45°, and m<1 is equal to m<2, so m<2 equals 45°

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3 years ago

Write 6 3/2 in surd form.

dolphi86 [110]

Answer:

$6\sqrt{6}$

Step-by-step explanation:

Simplify using $a\frac{n}{m} =\sqrt[m]{a^n} :\sqrt{6^3}$

Factor and rewrite the radicand in exponential form: $\sqrt{6^2 x 6}$

Rewrite the expression using $\sqrt[n]{ab} =\sqrt[n]{a}*\sqrt[n]{b} : \sqrt{6^2x\sqrt{6}$

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$6\sqrt{6}$

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2 years ago

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vodka [1.7K]

Answer:

624ft

Step-by-step explanation:

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4 0

3 years ago

Read 2 more answers

If it is warm outside, then I will not wear my jacket.

9966 [12]

If it's cold outside, then then I will wear my jacket.

Hope this helps.

4 0

4 years ago

"(b) use part (a) to find a power series for f(x) = 1 (4 + x)3 ."

Firdavs [7]

Assuming

$f(x)=\dfrac1{(4+x)^3}$

Recall that for $|x|$ ,

$\displaystyle\sum_{n\ge0}x^n=\frac1{1-x}$

Denote the above by $s(x)$ . Then

$s'(x)=\displaystyle\sum_{n\ge1}nx^{n-1}=\frac1{(1-x)^2}$
$s''(x)=\displaystyle\sum_{n\ge2}n(n-1)x^{n-2}=\frac2{(1-x)^3}$

Now,

$\dfrac1{(4+x)^3}=\dfrac1{4^3}\dfrac1{\left(1-\left(-\frac x4\right)\right)^3}=\dfrac1{2^7}\dfrac2{\left(1-\left(-\frac x4\right)\right)^3}$

which means we have

$f(x)=\dfrac1{2^7}s''(x)=\dfrac1{128}\displaystyle\sum_{n\ge2}n(n-1)\left(-\frac x4\right)^{n-2}$

which is valid for $\left|-\dfrac x4\right|$ , or $|x|$ .

4 0

3 years ago