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VARVARA [1.3K]
3 years ago
12

One-fourth of the girls and one-eighth of the boys in a class retake their school pictures. The photographer retakes pictures fo

r 16 girls and 7 boys. How many
students are in the class?
Mathematics
1 answer:
pychu [463]3 years ago
3 0
1/4 of the total girls = 16
1/4x = 16
x = 16 * 4
x = 64 girls 

1/8 of the total boys = 7
1/8x = 7
x = 7 * 8
x = 56 boys

for a total of (64 + 56) = 120 students <===
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Into a bucket, 3.705 liters of water flowed in 7.8 minutes. Enter the number of liters that flowed into the bucket each minute a
Anastaziya [24]
3.705 litres/7.8min
0.475 litres/min
5 0
3 years ago
For the following system, use the second equation to make a substitution for y in the first equation. 3x + y = 1 y + 4 = 5x What
Debora [2.8K]

Answer:

The answer is A.

Step-by-step explanation:

So we have the two equations:

3x+y=1\\y+4=5x

To make a substitution of the second equation into the first equation, we need to isolate the <em>y </em>variable in the second equation. Thus:

y+4=5x\\y=5x-4

Now, we can substitute this into the first equation. Therefore:

3x+y=1\\3x+(5x-4)=1\\3x+5x-4=1

7 0
3 years ago
Lisa made a shirt using 1/3 m of blue fabric and 3/5 m of red fabric how many meters of fabric did she use in all
Marysya12 [62]

Answer: 14/15 of a meter

Step-by-step explanation:

5 and 3 LCM is 15.

3/5 x 3 + 1/3 x 5= 9/15 + 5/ 15 = 14/15

5 0
3 years ago
39-50 find the limit.<br> 41. <img src="https://tex.z-dn.net/?f=%5Clim%20_%7Bt%20%5Crightarrow%200%7D%20%5Cfrac%7B%5Ctan%206%20t
Katyanochek1 [597]

Write tan in terms of sin and cos.

\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}

Recall that

\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1

Rewrite and expand the given limand as the product

\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)

Then using the known limit above, it follows that

\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}

4 0
2 years ago
FAST
sertanlavr [38]

Answer: The system consists of parallel lines

Step-by-step explanation:

Given system of lines :

y=\dfrac13x-4

3y-x=-7

Substitute y=\dfrac13x-4 in 3y-x=-7,  we get

3(\dfrac13x-4)-x=-7\\\\\Rightarrow\ x-12-x=-7\\\\\Rightarrow\ -12=-7 , which is not true.

That means , system has no solution.

i.e. they are representing parallel lines. [Parallel lines do not intersect and hence they do not have solution.]

6 0
3 years ago
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