If m<1 = 7x-19 and m<2 = x+5 find m<2

A pair of jeans was 30% off the original price. The sale resulted in a $24 discount.

Natalija [7]

Answer:

Step-by-step explanation:

We must consider the price of the Jeans as whole since it represents the full price of the jeans. The discount might be a limited only offer and does not represent the fullest price of the jeans.

To find the original price of the jeans we have two methods that in the end are going to lead to the same conclusion. The long method is to represent mathematically the information presented. Conceptually, we could interpret the statement saying that the original price minus 30% of the original price is equal to the original price minus the discount. Let us say the Original Price (OP)=x. then, mathematically we represent the statement as:

$x -0,3*x = x -24$

What we need to do now is to isolate x on one side of the equation. We have x as a common factor, so:

$x*(1-0,3)=x-24 \rightarrow 24+x*0,7=x$

$24=x-0,7*x \rightarrow 24 = x*(1-0,7)$

$24 = 0,3*x \rightarrow 24/0,3 =x \rightarrow x=80$

This one is the original price of the jeans: $80. There is a shortcut to this method, though. We know two things that are fundamental: the discount sale is $24 and the jeans are 30% off. These statements provide the same information; thus we can match them and say that:

$24 = 0,3*x \rightarrow 24/0,3 =x \rightarrow x=80$

And we find ourselves on the same conclusion.

Have a wonderful day :D

8 0

3 years ago

In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank

fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) $Amount \ of \, water \ remaining \ in \, the \ tank \ is \ \frac{x^3(6-\pi) }{6}$

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The volume of the sphere = $\frac{4}{3} \pi r^3$

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = $\frac{4}{3} \pi \frac{x^3}{8}$ = $\frac{1}{6} \pi x^3$

For tank B

Volume of tank = x³

The volume of the spheres = $8 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = $8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}$

For tank C

Volume of tank = x³

The volume of the spheres = $64 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = $64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}$

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The volume of the spheres = $512 \times \frac{4}{3} \pi r^3$

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = $512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}$

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = $x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}$

$Amount \ of \ water \, remaining \, in \, the \ tank = \frac{x^3(6-\pi) }{6}$ .

5 0

3 years ago

Highest common factor of 24 and 18

Alenkasestr [34]

6 is the highest common factor

8 0

4 years ago

Read 2 more answers

Help please I’m trying to do my homework and don’t understand

11Alexandr11 [23.1K]

Answer:

Answer is 9.

I think this is helpful. Vote it and like me

6 0

3 years ago

I’ve been trying to figure out this question on my homework and I’ve been confused for days now.

dsp73

Answer:

18m^12n^3

Hope this helps!!

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers