Math help with triangles please help (15 points)

there are 3/4 as many boys as girls in a class of fifth-graders if there are 35 students in the class how many a

Arlecino [84]

Alright so turn 3/4 into a decimal.
3/4 =.75

We need to set up an equation. To do that are need to give boys and girls their own variables.

Girls= X

And since there is 3/4 as many boys
Boys=. 75X

Now we know the sum of boys and girls equals 35
X+.75X=35

X+.75X=1.75X
Now we have 1.75X=35

Divide by 1.75 and get
20=X

Now plug 20 into the equations got from before.
Boys =. 75*X = 20 *. 75 = 15
Girls = X =20

Your Answer:
15 boys
20 girls

You can check
20+15=35
15/20=3/4

8 0

3 years ago

Read 2 more answers

if Hayden had 1224 golden apples and jenny had 1889 golden apple how many would the have all together

harina [27]

So, Hayden has 1224 Apples
Jenny has 1889 Apples.

All your doing is ADDING the apples to get your answer of : 3113 apples together.

4 0

3 years ago

If f(x) = - 4x + 8, what is the value of f(-4)

kompoz [17]

Answer:

Hi there!

Your answer is:

f(-4) = 24

Step-by-step explanation:

F(-4) means to plug in -4 for x!

-4(-4) +8

16+8= 24

f(-4) = 24

I hope this helps!

5 0

3 years ago

Read 2 more answers

Can you please answer the question?

Roman55 [17]

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that one side of the equality is equal to the expression of the other side, requiring the use of algebraic and trigonometric properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

The trigonometric expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the trigonometric expression $\sec \alpha \cdot \csc \alpha + 1$ .

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago

Add the two expressions. 3y+5 and y+24 im an idoit

Ostrovityanka [42]

No you aren't, just add the like terms! Meaning, add the terms with y and and the terms without y. You will get 4y+29

4 0

3 years ago