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3 years ago

A sum of money is divided in the ratio 2 : 3 : 7 .Given the largest share is $112,calculate the smallest share.

Mathematics

1 answer:

emmasim [6.3K]3 years ago

4 0

Answer:

$32

Step-by-step explanation:

A sum of money is divided in the ratio 2:3:7.

Let the share of each individual be represented by 2x,3x,7x respectively.

The largest share (7x) is given as $112.

$7x = 112$

$=>x = \frac{112}{7}$

$=>x = 16$

So the respective shares are $2\times16,3\times16,7\times16$

In other words, the shares are 32,48,112.

So the smallest share is $32

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Alex777 [14]

Answer:

It's actually right and scalene

Step-by-step explanation:

Since there is a ninety degree angle, it isn't obtuse or acute because the angle is a right angle. And since all the side lengths are different, it must be scalene in comparison to isoceles or equilateral.

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A city police force is made up of 24% females. If there are 72 females on the force, how many males are on the force?

erastovalidia [21]

Answer:

228 males

Step-by-step explanation:

Let x be the number of males.

This means x + 72 is the total number of people.

So $\frac{72}{x+72}$ is the fraction of females in the force.

Thus, we can form the equation:

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2 years ago

Pls show work! questions on pic

Lena [83]

Answer:

a.(10,23)

b.(7,30)

e.(20,20)

Step-by-step explanation:

We are given that

x=Width of garden

y=Length of garden

According to question

$x\geq 5$

$y\geq 20$

Perimeter of rectangle= $2(x+y)$

Where x=Width of rectangle

y=Length of rectangle

Fencing used = $2(x+y)$

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First we change inequality into equality

$x=5$ ...(1)

$y=20$ ...(2)

$2(x+y)=80$

$x+y=40$ ...(3)

Substitute x=0 in equation (3)

$y=40$

Substitute y=0 in equation (3)

$x=40$

Substitute x=0 and y=0 in equation (3)

$2(x+y)\leq 80$

$0\leq 80$

It is true. Therefore, the shaded region below the line.

$0\geq 5$

$0\geq 20$

These are false equation .Therefore, the shaded region above the line.

Substitute x=5 in equation (3)

$5+y=40$

$y=40-5=35$

Substitute y=20 in equation (3)

$x+20=40$

$x=40-20=20$

Intersection point of equation (1) and (3) is (5,35) and intersect point of equation (2) and (3) is (20,20).

Now,

a.(10,23)

Substitute in

$2(x+y)\leq 80$

$2(10+23)\leq 80$

$66\geq 80$

10>5 and 23>20

it satisfied all inequality.

Hence, it is a solution.

b.(7,30)

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it satisfied all inequality.

Hence, it is a solution.

c.(18,25)

18>5

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It does not satisfied third inequality

Hence, it is not a solution.

d.(8,35)

8>5

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Hence, it is not a solution.

e.(20,20)

$2(20+20)=80$

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