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3 years ago

Amy and Ben share £21 between them in the ratio 5 : 2. How much does Amy get?

Mathematics

1 answer:

Alika [10]3 years ago

6 0

Answer:

✑ $\underline{ \underline{ \sf{First ,\: Let's \: learn \: about \: the \: ratio}}} :$

Let us consider , there are 30 boys and 40 girls in a school. So, the ratio of number of boys to number of girls = $\sf{ \frac{30}{40} = \frac{3}{4} = 3:4}$ and read as 3 is to 4. We can say , the ratio compares two of more quantities of same unit. Until two quantities have same units , it is meaningless to compare by ratio. So, to find the ratio of two quantities , it is necessary to express them in same unit or of same kind. Since , they are in division form. So ,ratio is unit less quantity. Thus , the ratio is a comparison of two quantities of the same kind in division which doesn't have any units. For example : $\sf{ \frac{a}{b}}$ is a ratio and is read as ' a is to b ' in which ' a : is antecedent and ' b ' is called consequent.

✎ $\underline{ \underline{ \sf{Now \: let's \: solve : }}}$

☪ $\underline{ \sf{Solution}} :$

£ 21 is to be divided between Amy and Ben in the ratio 5 : 2. So, let Amy get £ 5x and Ben get £ 2x.

Then , According to the question , $\text{5x +2x = 21}$

Solve for x :

⇝ $\sf{7x = 21}$

⇝ $\sf{ \frac{7x}{7} = \frac{21}{7} }$

⇝ $\sf{x = 3}$

The value of x is 3. Now , substitute the value of x in 5x :

Amy received £ 5x = 5 × 3 = £ 15 .

☥ $\boxed{ \boxed{ \tt{Our \: final \: answer \: is \bold{£ \: 15}}}}$

Hope I helped ! ツ

Have a wonderful day / night ♡

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Calculate the volume of this rectangular object.

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Step-by-step explanation:

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4 years ago

I need some help with math

Irina18 [472]

Answer:

$(y-7)^2+(x-2)^2=16$

and

$(x+2)^2+(y-15)^2 = 9$

Step-by-step explanation:

The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where the coordinate (h,k) is the center of the circle.

Second Problem:

We can start with the second problem which uses this info very easily.
(h,k) in this problem is (-2,15) simply plug these into the equation. $(x--2)^2+(y-15)^2=r^2$ .
We can also add the radius 3 and square it so it becomes 9. The equation.
This simplifies to $(x+2)^2+(y-15)^2 = 9$ .

First Problem:

The first problem takes a different approach it is not in standard form. But we can convert it to standard form by completing the square.
$y^2-14y+x^2-4x+37=0$ first subtract 37 from both sides so the equation is now $y^2-14y+x^2-4x=-37$ .
$y^2-14y+x^2-4x+37=0$ by adding $(-\frac{b}{2a} )^2$ to both the x and y portions of this equation you can complete the squares. $(-\frac{b}{2a})^2=(-\frac{-14}{2(1)})^2$ and $(-\frac{-4}{2(1)})^2$ which equals 49 and 4.
Add 49 and 4 to both sides and the equation is now: $y^2-14y+49+x^2-4x+4=-37+49+4$ You can simplify the y and x portions of the equations into the perfect squares or factored form $(y-7)^2$ and $(x-2)^2$ .
Finally put the whole thing together. $(y-7)^2+(x-2)^2=16$ .