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3 years ago

In a group of 80 children there are 3 times as many boys as girls how many girls are there

Mathematics

1 answer:

insens350 [35]3 years ago

7 0

80=x+3x
80=4x
20=x

There are 20 girls.

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PLEASE HELP DO QUESTION 5 FOR BRAINLEYIST AWNSER.

zavuch27 [327]

There is no question 5?

5 0

3 years ago

Draw a rectangular fraction model to explain your thinking. Then, write a multiplication sentence. A. 2 3 of = 3 5

myrzilka [38]

Answer:

$\frac{2}{3}\ of\ \frac{3}{5} = \frac{2}{5}$

Step-by-step explanation:

Given

$\frac{2}{3}\ of\ \frac{3}{5}$

Solving (a): Multiplication sentence

$\frac{2}{3}\ of\ \frac{3}{5}$

Rewrite of as *

$\frac{2}{3}\ of\ \frac{3}{5} = \frac{2}{3}\ *\ \frac{3}{5}$

Solve

$\frac{2}{3}\ of\ \frac{3}{5} = \frac{2}{5}$

Hence, the multiplication sentence is:

$\frac{2}{3}\ of\ \frac{3}{5} = \frac{2}{5}$

Solving (b): Rectangular fraction model

In (a) above, the result is 2/5

The fraction model will be represented as thus:

Draw a rectangle
Partition in to 5 equal parts (5 represents the denominator)
Shade 2 of the 5 partitions (2 represents the numerator)

<em>See attachment for model</em>

5 0

3 years ago

Please, I need help with my homework. I not asking anybody to do it, but I need help for it.

jenyasd209 [6]

Answer:

1) x = -2

Step-by-step explanation:

1) x + 7 + x + 11 = 14

2x + 18 = 14

2x = 14 - 18

2x = -4

x = -4/2 = -2

4 0

3 years ago

Find the inverse of g(x)=1/3x - 7 and then find its domain and range if you can show steps that would be gladly appreciated than

Schach [20]

Answer:

g-1(x) = 3(x + 7).

Domain is All Real values of x.

Range is All Real values of the function.

Step-by-step explanation:

Let y = 1/3 x - 7

1/3 x = y + 7

Multiply both sides by 3:

x = 3(y + 7)

So the inverse of g(x) = g-1(x) = 3(x + 7).

Domain is All Real values of x.

Range is All Real values of the function.

6 0

3 years ago

Can you guys help please im so confused

marissa [1.9K]

Answer:

a) see below

b) 40x20 meters

Step-by-step explanation:

Write down what you know:

- The area of the enclosure is length*width, so

$A = x \cdot y$

- The length of the fencing is 80 meters, so

$2y + x = 80$

Now we have to combine these two equations above, and get rid of y in the process.

First rewrite the second as:

$y = 40 - \frac12 x$

Then substitute for y in the first:

$A = x (40 - \frac12 x) = 40x - \frac12x^2$

b) To maximize A, find the zero of the first derivative:

$A'(x) = 40 - x = 0 \implies x=40$

So y = (80-40)/2 = 20 meters.

8 0

2 years ago