All categories

2 years ago

Share £117 In the ratio 5:2:6

Mathematics

1 answer:

julia-pushkina [17]2 years ago

7 0

6.24 hope this is the answer

You might be interested in

A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.

Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}$

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}$

6 0

2 years ago

The table below shows the amount of lemon juice and sugar needed to make three different-sized batches of lemonade using the sam

mash [69]

Answer:

j = 2.5s

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Lemon juice (mL) Sugar (gr)

Batch A 500 200

Batch B 750 300

Batch C 1,500 600

2. Write an equation to describe the relationship between j, the amount of lemon juice in mL, and s, the amount of sugar in g.

Ratio = Amount of lemon juice/Amount of sugar

Ratio = 500/200 = 750/300 = 1,500/600

Ratio = 2.5

Now, we can write the equation, this way:

j = 2.5s

8 0

2 years ago

Read 2 more answers

It is known that a cable with a cross-sectional area of 0.300.30 sq in. has a capacity to hold 2500 lb. If the capacity of the

vesna_86 [32]

Answer:

0.84 square in

Step-by-step explanation:

Since the capacity of the cable is proportional to its cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold

2500 / 0.3 = 8333.33 lb/in

To old 7000 lb it the cross-sectional area would need to be

7000 / 8333.33 = 0.84 square in

5 0

2 years ago

Read 2 more answers

A sports company wants to package a ball with a 1.5-inch radius in sets of two. They have two options: a cylinder or a square pr

guajiro [1.7K]

Step 1

Find the Volume of the two balls

we know that

the Volume of a sphere is equal to

$V=\frac{4}{3} \pi r^{3}$

In this problem we have

$r=1.5\ in$

Substitute

$V=\frac{4}{3} \pi 1.5^{3}=4.5\pi\ in^{3}$

the volume of the two balls is equal

$V=2*4.5\pi=9 \pi\ in^{3}$

Step 2

Find the volume of a cylinder

we know that

The Volume of a cylinder is equal to

$V=\pi r^{2}h$

In this problem we have

$r=1.5\ in$

$h=4*r=4*1.5=6\ in$

Substitute

$V=\pi (1.5^{2})(6)=13.5 \pi\ in^{3}$

Step 3

Find the amount of wasted space in the cylinder

Subtract the volume of the two balls from the volume of the cylinder

$13.5\pi\ in^{3}-9\pi\ in^{3}=4.5\pi\ in^{3}=14.14\ in^{3}$

Step 4

Find the volume of the square prism

The volume of the prism is equal to

$V=Bh$

where

B is the area of the base of the prism

h is the height of the prism

In this problem we have

$B=(2*1.5)^{2}=9\ in^{2}$ -----> the base is equal to the diameter

$h=4*r=4*1.5=6\ in$

Substitute

$V=9*6=54\ in^{3}$

Step 5

Find the amount of wasted space in the prism

Subtract the volume of the two balls from the volume of the prism

$54\ in^{3}-9\pi\ in^{3}=25.73\ in^{3}$

Step 6

Compare the amount of wasted space

the amount of wasted space in the cylinder is $14.14\ in^{3}$

the amount of wasted space in the prism is $25.73\ in^{3}$

Find the difference

$25.73\ in^{3}-14.14\ in^{3}=11.59\ in^{3}=11.6\ in^{3}$

The package that has the least amount of wasted space is the cylinder

therefore

the answer is the option

the cylinder because it has approximately 11.6 in.3 less wasted space than the prism

5 0

2 years ago

Read 2 more answers

John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21 About how far does

Agata [3.3K]

Answer:

John ski down the mountain is 1285.37 feet.

Step-by-step explanation:

Given : John is skiing on a mountain with an altitude of 1200 feet. The angle of depression is 21.

To find : About how far does John ski down the mountain ?

Solution :

We draw a rough image of the question for easier understanding.

Refer the attached figure below.

According to question,

Let AB be the height of mountain i.e. AB=1200 feet

The angle of depression is 21 i.e. $\theta=21^\circ$

We have to find how far does John ski down the mountain i.e. AC = ?

Using trigonometric,

$\cos\theta = \frac{AB}{AC}$

$\cos(21)= \frac{1200}{AC}$

$AC=\frac{1200}{\cos(21)}$

$AC=1285.37$

Therefore, John ski down the mountain is 1285.37 feet.

7 0

2 years ago