Ask question

Ask question

All categories

2 years ago

10

There are 12 girls in the swim team. they make up 48% of he students on the team. How many boys are on the swim team?

2 answers:

zaharov [31]2 years ago

8 0

Answer:

b

Step-by-step explanation:

13

Hunter-Best [27]2 years ago

6 0

Answer:

25

Step-by-step explanation:

12/x = 48/100

48/100 = 0.48

12 = 0.48x

12/0.48 = 25

You might be interested in

Consider the accompanying matrix as the augmented matrix of a linear system. State in words the next two elementary row operatio

Lubov Fominskaja [6]

Answer:

The answer is below

Step-by-step explanation:

The next step to solve the system, would be to divide the second row by 3 and we would be left with: 3 * R2

[1 -5 3 0 -2

0 1 -7/3 0 4/3

0 0 1 2 -2]

Then what we will do is multiply row 3 by 7/3 and then subtract it from row 2, that is, R2 - 7/3 * R3, and it would look like this:

[1 -5 3 0 -2

0 1 0 14/3 -10/3

0 0 1 2 -2]

And these would be the next two steps in the process of solving the system.

4 0

3 years ago

Can u help me solve this pls

lisov135 [29]

Answer:

First option: (Substitute)

2x+3 (x+1.5) = 12

3 0

2 years ago

Read 2 more answers

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

How much does Craig earn each month,in dollars?

allsm [11]

Answer:

Craig must be rich LOL cuz It says its 1/20 of his paycheck each month so u have to multiply by 20

Step-by-step explanation:

720*20 = 14440

7 0

2 years ago

How to turn pounds to ounces

aalyn [17]

1 pound = 16 ounces
multiply the amount of pounds you have by 16 (that's how many ounces is in a pound) your answer should be whatever you get
example: you have 13 pounds, just multiply it by 16, the answer is 208.
so there are 208 ounces in 13 pounds.
or you could go google and ask them.
hope this helps.

8 0

3 years ago