Ask question

Ask question

All categories

3 years ago

12

A day at Camp there were 56 girls and 42 boys the campers need to be split into equal groups each has either all girls are all b

oys what is the greatest number of campers each can have

1 answer:

Olegator [25]3 years ago

3 0

number of girls = 56

number of boys = 42

highest common factor = 14

greatest number of campers per group is 14

You might be interested in

Mr. Liaw's ceramic mug has a volume of 60, but a radius of 2 inches the ceramic mug?

Anuta_ua [19.1K]

Answer:

<u>4.77 in.</u>

Step-by-step explanation:

Height = volume / πr²

= 60 / 3.14 x 4

= 15/3.14

= <u>4.77 in.</u>

4 0

2 years ago

Read 2 more answers

$299 for 4 tires this is unit rate i don't get this please help me

arsen [322]

Well first I think you need to divided <span />

6 0

3 years ago

Read 2 more answers

The midpoint of (-4,15) and (22,3)

ahrayia [7]

Answer:

(9, 9)

Step-by-step explanation:

midpoint of (-4,15) and (22,3)

$= \bigg( \frac{ - 4 + 22}{2}, \: \: \frac{15 + 3}{2} \bigg) \\ \\ = \bigg( \frac{18}{2}, \: \: \frac{18}{2} \bigg) \\ \\ = ( 9, \: \: 9) \\ \\$

8 0

3 years ago

Plz answer and provide an explanation

riadik2000 [5.3K]

I believe it is A and Yes

The bigger the price is on an item, the more tax you would have to pay, so both numbers increase. As for whether or not there is a line through the dots, you use a line when you could go inbetween the points. For example, with this question you can have 0.24, 0.86, or any other number as the price and it would make sense. If the question was sales tax on a certain number of shirts, well, you can't buy half a shirt so you don't need anywhere inbetween the dots.

Proportional in these kinds of problems means whether or not the line curves. If it doesn't and is a straight line, then it is proportional.

8 0

2 years ago

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

5 0

3 years ago