Solve for x. 0.9(x+1.4)−2.3+0.1x=1.6

On the Venn diagram, which region(s) represent the union of Set A and Set B (A⋃B)?

slega [8]

The set A U B is represented by shading both circles A and B (also the overlapping region as well). Any point in set A U B will be in either A, B or both.

7 0

3 years ago

Read 2 more answers

An elevator can hold a maximum of 1500 pounds. An average child weighs 75 pounds and the average adult weighs 150 pounds. The el

CaHeK987 [17]

Let's take this problem step-by-step:

Let's first set up some variables:

c: # of children
a: # of adults

Let's examine the information given:

Elevator can hold a maximum of 1500 pounds

⇒ average child is 75 pounds

⇒ average adult is 150 pounds

⇒ therefore: $75c + 150a\leq 1500$

Elevator can fit no more than 14 people

⇒ therefore: $c + a \leq 14$

Let's graph the equations:

$75c+150a\leq 1500\\c+a\leq 14$

⇒ look at the image attached

The point at which the two graphs intersect:

⇒ is the solution that represents the amount of children and adults and

their combine weight

With the horizontal axis being the # of children and vertical axis being the # of adults:

⇒ the solution is 8 children and 6 adults

Answer: 8 children and 6 adults

Hope that helped!

7 0

1 year ago

What decimal is equal to 3/10

Aleks [24]

Answer:

0.3

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

An abstract sculpture in the form of a triangle has a base 21 meters and a height of 12 meters. Find the are of the triangle.

charle [14.2K]

The area of a triangle could be determined using the following formula
$\boxed{a= \frac{1}{2} \times b \times h}$

plug in the numbers
$a = \frac{1}{2} \times 21 \times 12$
$a = \frac{1}{2} \times 252$
$a= \frac{252}{2}$
a = 126

The area of the triangle is 126 square meters

8 0

3 years ago