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3 years ago

12

✨SOLVE THE SOLUTION TO EACH SYSTEM OF INEQUALITIES ✨ PLS HELPPPPP ME

1 answer:

muminat3 years ago

7 0

i saw its C because where the solutions are located. sorry if it’s wrong, i’m rusty

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If 5+ 20 -22-3х - 10.2-2x+5, what is the value of х?

Elodia [21]

Answer:

<h2>3</h2>

Step-by-step explanation:

$5+20\cdot2^{2-3x}=10\cdot2^{-2x}+5\qquad\text{subtract 5 from both sides}\\\\20\cdot2^{2-3x}=10\cdot2^{-2x}\qquad\text{divide both sides by 10}\\\\2\cdot2^{2-3x}=2^{-2x}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\2^{1+2-3x}=2^{-2x}\\\\2^{3-3x}=2^{-2x}\iff3-3x=-2x\qquad\text{subteact 3 from both sides}\\\\-3x=-2x-3\qquad\tex\qquad\text{add}\ 2x\ \text{to both sides}\\\\-x=-3\qquad\text{change the signs}\\\\x=3$

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3 years ago

Jonny has five Apples he gives four to Sarah how many apples do he have left ?

Korolek [52]

Answer:

He has one.

Step-by-step explanation:

He has five, and gives sarah four. That means that you take all but one away from 5. 5-4=1.

6 0

4 years ago

Help please is for now

Viefleur [7K]

Answer:

15

Step-by-step explanation:

4 0

3 years ago

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A worker on a scaffolding 75 ft above the ground needs to lift a 500 lb bucket of cement from the ground to a point 30 ft above

Lemur [1.5K]

Answer:

W = 15900 ft.lb

Step-by-step explanation:

From the given information.

Consider y to be the distance from the ground towards the bucket of the cement.

At height y, suppose the bucket is lifted by Δy.

Then, the workdone can be = 500Δy + 0.5(75 - y) Δy

where;

The bucket of the cement = 500Δy

The remaining of the rope = 0.5(75 - y) Δy

However; the workdone (W) can be calculated as:

$W = \int^{30}_{0} \ 500 \ dy + \int^{30}_{0} \ 0.5 (75-y ) \ dy$

$W = 500 \times 30 + 0.5 \bigg(75 \times 30 - \dfrac{1}{2} \times 30^2 \bigg )$

W = 15000 + 0.5 ( 2250 - 450)

W = 15000 + 900

W = 15900 ft.lb

6 0

3 years ago

A car travels 3 hours at 50mph; then travels

kipiarov [429]

Answer:

430

3x50 is 150 plus 280= 430

3 0

4 years ago

Read 2 more answers