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3 years ago

There are 42 runners in a race. How many different ways can the runners finish first, second, and third?

Mathematics

1 answer:

Murljashka [212]3 years ago

4 0

sorry wrong thing yeah but im not sure

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This is number nine if you wanted to help please

andrew-mc [135]

Answer: 60 square units

Step-by-step explanation:

If you're supposed to find area it would be 5x7 + 5x5 = 35 + 25 = 60

5 0

2 years ago

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In order for you to answer the question correctly, please use the following image down below:

adell [148]

Answer:

Step-by-step explanation:

24^2 = (12)(12+x)

576 = 144 + 12x

432 = 12x

36 = x

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2 years ago

Absolute value inequality

Marizza181 [45]

|x| because just Google it if I'm wrong

8 0

3 years ago

PLSS HELP

Dmitry [639]

Answer:

$8,000

Step-by-step explanation:

Let the store earned $x in December.

Therefore,

Money spent to buy new inventory $=\frac{1}{4} x$

Remaining money $= x - \frac{1}{4} x =\frac{3}{4} x$

Money used to pay bills $=\frac{1}{2} \times \frac{3}{4} x=\frac{3}{8} x$

Money still left over = $3,000

Total money earned in December $= \frac{1}{4} x+ \frac{3}{8} x+3,000$

$\therefore x= \frac{1}{4} x+ \frac{3}{8} x+3,000$

$\therefore x= \frac{2}{8} x+ \frac{3}{8} x+3,000$

$\therefore x= \frac{5}{8} x+ 3,000$

$\therefore x- \frac{5}{8} x=3,000$

$\therefore \frac{8x-5x}{8} =3,000$

$\therefore \frac{3x}{8} =3,000$

$\therefore x =3,000\times \frac {8}{3}$

$\therefore x =1,000\times 8$

$\therefore x =\$8,000$

Thus, total money earned in December is $8,000.

3 0

2 years ago

Calculus: Help ASAP

wariber [46]

$\bf \displaystyle \int~\cfrac{sec^2(x)}{\sqrt{1+tan(x)}}\cdot dx\\\\
-------------------------------\\\\
u=1+tan(x)\implies \cfrac{du}{dx}=sec^2(x)\implies \cfrac{du}{sec^2(x)}=dx\\\\
-------------------------------\\\\
\displaystyle \int~\cfrac{\underline{sec^2(x)}}{\sqrt{u}}\cdot\cfrac{du}{\underline{sec^2(x)}}\implies \int~\cfrac{1}{\sqrt{u}}\cdot du\implies \int~u^{-\frac{1}{2}}\cdot du
\\\\\\
2u^{\frac{1}{2}}\implies 2\sqrt{1+tan(x)}+C$

6 0

3 years ago

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