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3 years ago

There are 16 girls in a school club. the number of girls is four more than twice the number of boys. find the number of boys.

1 answer:

3 0

The number of boys . . . . . B
Twice the number of boys . . . 2B
4 more than twice the number of boys . . . 2B+4

You said that the number of girls is equal to that, so . . . G = 2B+4

But you also told us what G is: 16 = 2B+4

Subtract 4 from each side: 12 = 2B

Divide each side by 2 : 6 = B

There you have it. There are 6 boys in the club.

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3 years ago

A rectangle is 6yards longer than it is wide. Find the dimensions of the rectangle if it’s area is 187 square yards

emmasim [6.3K]

Answer:

Width = 11 yards

Length = 17 yards

Step-by-step explanation:

First of all, the length of the rectangle is 6 yards longer than the width, this means, length = width + 6 yards. This dimensions can be represented on figure 1, where <em>w</em> is width, and <em>l</em>, for length.

We know the area of a rectangle is A = width x length

For our case 187 = w . (w + 6)

Using the Distributive Property for the multiplication we obtain

$187 = w^{2} +6w$

$w^{2} +6w-187 =0,$

Using the quadratic formula $w=\frac{-b\±\sqrt{b^{2}-4ac } }{2a}$ where a = 1, b = 6, c = - 187 and replacing into the formula, we will have:

$w=\frac{-6\±\sqrt{6^2-4(1)(-187)} }{2(1)}$

$w=\frac{-6\±\sqrt{36+748} }{2}=\frac{-6\±\sqrt{784} }{2}=\frac{-6\±28}{2}$

We have two options: $w=\frac{-6+28}{2}=\frac{22}{2}=11 yards$

$w=\frac{-6-28}{2}=\frac{-34}{2}=-17 yards$ But a distance (width) can not be negative so, this answer for w must be discarded.

The answer must be width = 11 yards.

To find the length $l =\frac{187}{11}=17 yards$

6 0

2 years ago

Simplify the expression 3/2 (5/3) -1 1/4 + 1/2 Given the answer as a simplified fraction. The simplified expression is a/b.

ella [17]

<h2>Hello!</h2>

The answer is:

The simplified fraction is:

$\frac{7}{4}$

To solve this problem we must remember the following:

- Addition or subtraction of fractions, we add or subtract fractions by the following way:

$\frac{a}{b}(+-)\frac{c}{d}=\frac{ad(+-)bc}{bd}$

- Product of fractions, the multiplication of fraction is linear, meaning that we should multiply the numerator by the numerator and denominator by denominator, so:

$\frac{a}{b}*\frac{c}{d}=\frac{ac}{bd}$

- Convert mixed number to fraction,

$a\frac{b}{c}=a+\frac{b}{c}=\frac{ac+b}{c}$

So, solving we have:

$\frac{3}{2}*\frac{5}{3}-1\frac{1}{4}+\frac{1}{2}=\frac{3*5}{2*3} -1\frac{1}{4}+\frac{1}{2}\\\\\frac{3*5}{2*3} -1\frac{1}{4}+\frac{1}{2}=\frac{15}{6} -1\frac{1}{4}+\frac{1}{2}\\\\\frac{15}{6} -1\frac{1}{4}+\frac{1}{2}=\frac{15}{6} -(1+\frac{1}{4})+\frac{1}{2}\\\\\frac{15}{6} -(1+\frac{1}{4})+\frac{1}{2}=\frac{15}{6}-(\frac{4+1}{1*4})+\frac{1}{2}\\\\\frac{15}{6}-(\frac{4+1}{4})+\frac{1}{2}=\frac{15}{6}-\frac{5}{4}+\frac{1}{2}$

$\frac{15}{6}-\frac{5}{4}+\frac{1}{2}=(\frac{5}{2}-\frac{5}{4})+\frac{1}{2}\\\\(\frac{5}{2}-\frac{5}{4})+\frac{1}{2}=(\frac{(4*5)-(2*5)}{8})+\frac{1}{2}\\\\(\frac{(4*5)-(2*5)}{8})+\frac{1}{2}=(\frac{20-10}{8})+\frac{1}{2}\\\\(\frac{20-10}{8})+\frac{1}{2}=(\frac{10}{8})+\frac{1}{2}\\\\(\frac{10}{8})+\frac{1}{2}=\frac{5}{4}+\frac{1}{2}\\\\\frac{5}{4}+\frac{1}{2}=\frac{(5*2)+(4*1)}{2*4}\\\\\frac{(5*2)+(4*1)}{2*4}=\frac{10+4}{8}=\frac{14}{8}\\\\\frac{14}{8}=\frac{7}{4}$