All categories

3 years ago

Consider the inequality X > - 2.5 Determine whether each value of X makes this inequality true Select ALL that apply.

Mathematics

2 answers:

Leno4ka [110]3 years ago

6 0

Answer:

The correct answers are: 0.5, 3, and 10.

Basically, all numbers that are greater than -2.5 can make this inequality true.

Let me know if this helps!

grigory [225]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

Elijah’s workout at the gym consist of the elliptical and treadmill. After burning 450 cal on the elliptical machine, Elijah swi

krek1111 [17]

The answer is 375

30*12.5 is equal to 15*25
15*25= 375
Plz mark as Brainly

7 0

2 years ago

Given that the lines are parallel, what would be the value of angle "a" in the diagram found below?

yawa3891 [41]

Cannot be determined as there isn’t an angle

8 0

3 years ago

Read 2 more answers

Help for brainly and poibts

lana [24]

It would be more than one because 3/5 + 3/4 equals 1 7/20

3 0

3 years ago

Read 2 more answers

What is the area of the figure below

jek_recluse [69]

Answer:

Option A

Step-by-step explanation:

Area of the composite figure = Area of ΔABH + Area of BCGH + Area of ΔDEF

Area of ΔABH = $\frac{1}{2}(\text{Base})(\text{Height})$

= $\frac{1}{2}(6)(8-5)$

= 9 in²

Area of the trapezoid = $\frac{1}{2}(CG+BH)(\text{Distance between these lines})$

= $\frac{1}{2}(6+4)(5)$

= 25 in²

Area of ΔDEF = $\frac{1}{2}(8)(6)$

= 24 in²

Therefore, area of the given figure = 9 + 25 + 24

= 58 in²

Option A will be the correct option.

3 0

2 years ago

Approximate the change in the volume of a sphere when its radius changes from r = 40 ft to r equals 40.05 ft (Upper V (r )equal

alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere ( $V$ ), measured in cubic feet, is represented by the following formula:

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

Where $r$ is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

$\Delta V = \frac{\partial V}{\partial r}\Delta r$

The derivative of the volume as a function of radius is:

$\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}$

Then, the change in volume is expanded:

$\Delta V = 4\pi \cdot r^{2}\cdot \Delta r$

If $r = 40\,ft$ and $\Delta r = 40\,ft-40.05\,ft = 0.05\,ft$ , the change in the volume of the sphere is approximately:

$\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)$

$\Delta V \approx 1005.310\,ft^{3}$

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

7 0

3 years ago