Ask question

Ask question

All categories

3 years ago

9

Two-thirds of the students in Harry’s math class and three-fifths of the students in Amani’s math class are boys. Which question

about the classes is best modeled with a division expression?
A.If there are 12 boys, how many students are in Harry’s class?
B.If there are 24 students, how many boys are in Amani’s class?
C.What fraction of the students in Harry’s class are girls?
D.What fraction of the students in Amani’s class are honors students if the fraction is equal to mc013-1.jpg the fraction of students who are boys?

2 answers:

nata0808 [166]3 years ago

8 0

There are students in Harry's class. The fraction of girls in Harry's class is 1/3.

marissa [1.9K]3 years ago

5 0

Answer:

its c

Step-by-step explanation:

You might be interested in

Find the coordinates of the midpoint of a segment with the given

galben [10]

Answer:

(9,5.5)

Step-by-step explanation:

6 0

3 years ago

Which type of parent function is f(x) =^3cubed x

sergij07 [2.7K]

Answer:

Cube Root

Step-by-step explanation:

Because there is a 3 outside the symbol.

That means the equation is stating: y * y * y = x

Hope this helps :)

7 0

2 years ago

What is the equation of the line that has a slope of 1/5 and a y-intercept of -1

DiKsa [7]

Answer:

C y = 1/5x - 1

Step-by-step explanation:

Slope intercept form: y = mx + b where m = slope and b = y-intercept.

Substitute into the slope intercept form m= 1/5 and b = -1

y = 1/5x - 1

4 0

2 years ago

63.625 rounded to the nearest tenth

Ket [755]

Answer:

63.6

Step-by-step explanation:

63.625

63.6

7 0

2 years ago

Read 2 more answers

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

$x=(\frac{y+3}{2})^2$

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

Read 2 more answers