30% of number A is 10 more than 20% of number B. 30% of number B is 35 more than 20% of number A. Find the numbers A and B.

Mathematics

1 answer:

anygoal [31]3 years ago

5 0

Answer:

A = 76.85

B =65.28

Step-by-step explanation:

(30/100)A = 10 + (20/100)B

0.3A - 0.2B = 10 ...... equation (i)

(30/100)B + 35 = (20/100)A

0.2A - 0.3B = 35 ........ equation (ii)

From equation (i)

0.3A = 10 - 0.2B

A = (10 - 0.2b) / 0.3

A = 33.33 - 0.67B ........equation (iii)

Put equation (iii) into equation (ii)

0.2(33.33 - 0.67B) - 0.3B = 35

6.67 - 0.134B - 0.3B = 35

0.434B = 35 - 6.67

B = 28.33 / 0.434

B = 65. 275 = 65.28

Put B = 65.28 into equation (i)

0.3A - 0.2B = 10

0.3A - 0.2(65.28) = 10

0.3A - 13.056 = 10

0.3A = 10 + 13.056

A = 23.056/ 0.3

A = 76.85

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The curve y = |x|/ 5 − x2 is called a bullet-nose curve. find an equation of the tangent line to this curve at the point (2, 2).

diamong [38]

We have the following curve:

$y=\frac{\left | x \right |}{5-x^2}$

So we need to find an equation of the tangent line to this curve at the point $(2,2)$ . So let's find out if this point, in fact, belongs to the curve:

 $If \ x=2 \\ \\ y=\frac{\left | 2 \right |}{5-2^2}=\frac{\left | 2 \right |}{5-4}=2$ .

We also know that:

 $\left | x \right |= \left \{ {{x \ if \ x \ \geq 0} \atop {-x \ if \ x\ \textless \ 0}} \right.$ 

Given that the point is:

 $(2,2)\ that \ is: \\ x=2\ \textgreater \ 0$

Then we will say that:

$\left | x \right |=x$

Therefore:

$y=\frac{x}{5-x^2}$

Computing the derivative:

$y=\frac{x}{5-x^2} \\ \frac{dy}{dx}= \frac{(1)(5-x^2)-(-2x)x}{(5-x^2)^2}=\frac{5+x^2}{(5-x^2)^2}$

So the derivative solved for $x=2$ is in fact the slope of the line at the point $(2,2)$ , then:

$\frac{dy}{dx} |_{x=2}=\frac{5+x^2}{(5-x^2)^2}|_{x=2}=\frac{5+2^2}{(5-2^2)^2}=9 \\ \\ \therefore m=\frac{dy}{dx} |_{x=2}=9$

Finally, the tangent line is:

$y-y_1=m(x-x_1) \\ \therefore y-2=9(x-2) \\ \therefore \boxed{y=9x-16}$

This is shown in the figure below.