All categories

3 years ago

there are 12 students working in the library if 3/4 of them are girls how many girls are working in the library

2 answers:

6 0

If you would like to know how many girls are working in the library, you can calculate this using the following steps:

3/4 of 12 students = 3/4 * 12 = 3 * 3 = 9 girls

Result: There are 9 girls working in the library.

navik [9.2K]3 years ago

5 0

So 3/4 of 12
remember that 12=12/1
and if you ahve x/y times z/t that equals to (xz)/(yt) so
'of' means multiply
3/4 times 12/1=(3 times 12)/(4 times 1)=36/4=9/1=9
9 are working in the library

You might be interested in

The perimeter of a rectangle equals one and a half times its area. Express the length of the rectangle in terms of the width (us

WITCHER [35]

Answer:

Step-by-step explanation:

let x and y be length and width of rectangle.

Perimeter=2(x+y)

area=xy

2(x+y)=1/2 xy

4(x+y)=xy

4x+4y=xy

x y-4y=4 x

y(x-4)=4 x

$y=\frac{4x}{x-4}$

4 0

3 years ago

PLSS HELP ME WITH THIS! 15 POINTS!

irina [24]

Answer:

2w - 13

Step-by-step explanation:

here we just simplify the expression,

3w - 4 - w - 9

becomes:

2w - 13

5 0

2 years ago

Read 2 more answers

Which of the following is the correct factorization of the polynomial below?

alexandr1967 [171]

X + 7 might be right

8 0

3 years ago

WHAT IS 15.7188888888 X 18 TO THE POWER OF SIX DIVIDED BY 7 Plus 3

lakkis [162]

I think it is green

4 0

3 years ago

Read 2 more answers

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago