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3 years ago

What are the solutions?

Mathematics

2 answers:

777dan777 [17]3 years ago

8 0

Answer:

I cant really see the answers try taking another picture

Step-by-step explanation:

Alexeev081 [22]3 years ago

5 0

Take another pic please

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-8<2x-4<4 solve the inequality

il63 [147K]

Answer:

Step-by-step explanation:

The inequality will be split into two

It is know that, if a<b<c

Then a<b and b<c

-8<2x-4<4

Apply that to this

Then,

-8<2x-4. Equation 1

Also,

2x-4<4 equation 2

Solving equation 1

-8<2x-4

Add 4 to both side of the equation

-8+4<2x-4+4

-4<2x

Divide both sides by 2

-4/2<2x/2

-2<x

Note, if a is less than b, then, b is greater than a, e.g. 4 is less than 10, this implies 10 is greater than 4

Therefore,

-2<x

Then, x greater than -2

Equation 2

2x-4<4

Add 4 to both side of the inequalities

2x-4+4<4+4

2x<8

Divide both side by 2

Then,

2x/2<8/2

x<4

Therefore x is between -2 and +4.

Check attachment for graphical solution

7 0

3 years ago

Grain is falling from a chute onto the ground, forming a conical pile whose diameter is always three times its height. how high

bezimeni [28]

Given that grain is falling from a chute onto the ground, forming a conical pile whose diameter is always three times its height.

So if D is the diameter and h is the height of the conical pile then we can write:

D=3h

We know that diameter = 2r, where r is the radius

2r=3h

$r=\frac{3h}{2}$

Volume of conical pile is given by formula

$V=\frac{1}{3}\pi r^2h$

Given that volume is 1110 cubic feet.

Now plug the values of Volume and r into equation of volume

$1110=\frac{1}{3}\pi (\frac{3h}{2})^2h$

$1110=\frac{1}{3}\pi\cdot\frac{9h^2}{4}\cdot h$

$1110=\pi\cdot\frac{3h^3}{4}$

$1110\cdot\frac{4}{3\pi}=h^3$

$471.098631552=h^3$

take cube root of both sides

7.78103342467=h

Hence height is approx 7.78 feet.

7 0

3 years ago

IF YOU ANSWER RIGHT I MIGHT GIVE BRAINLIST!!!

Pie

Answer:

1. Positive, 1+2=3

2. Negative, -1-2=-3

Step-by-step explanation:

If you look at both in a graphing perspective, the point (1,2) is in Quadrant I. likewise, adding 2 to the x-coordinate will also result in the point (3,2), also in Quadrant I, where the x coordinate is positive. The point (-1,2) is in Quadrant II, and adding -2 to the x coordinate keeps it in Quadrant II, where the x-coordinate is negative.

8 0

3 years ago

Read 2 more answers

ASAP someone please simply two exponents

zhannawk [14.2K]

Answer:

1.) $m^{15}$