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slega [8]
3 years ago
5

Solve the following inequalities and give the solutions in both set and interval notations.

Mathematics
1 answer:
nikdorinn [45]3 years ago
5 0

4y + 3 > 23          subtract 3 from both sides

4y > 20                divide both sides by 4

y > 5

 

-2y > -2                divide both sides by -2 (the order changes)

y < 1

 

(-oo, 1) ∪ (5, oo)

Upvote

SO SORRY IF THESE IS WRONG:(( JUST SAY IT TO ME NOT TO BULLY ME

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Pls help me........ I tried lol
diamong [38]

its step 2 he did not subtract it right hope this helps and have a great day and get some sleep lol

7 0
2 years ago
Read 2 more answers
Evaluate the following expression.<br> 6 -3
Vinvika [58]

Answer:

3

Step-by-step explanation:


8 0
3 years ago
Whats the figures and how do I find the area?​
iragen [17]

Answer:

This is trapezium.

Area of this figure =1/2h(d1-d2)

1/2×5(14-8)

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8 0
3 years ago
(I've been trying to figure this out for 3 days and I really need help)
liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

1)

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7

2)

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55

3)

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

4)

pretty much the same thing, we get the volume of the cone and its top, add them up.

\bf \stackrel{\textit{cone's volume}}{\cfrac{\pi (3)^2(8)}{3}}~~~~+~~~~\stackrel{\stackrel{\textit{half a sphere}}{\textit{top's volume}}}{\cfrac{4\pi 3^3}{3}\div 2}\implies 24\pi +18\pi \implies 42\pi ~~\approx~~131.95~in^

8 0
3 years ago
The density of glass is 2.6 g/cm3 and the density of lead is 11.3 g/cm3. Does this mean there are more particles of lead in 1 cm
Colt1911 [192]
The answer is that there are more particles in 1 cubic cm. this is because within 1 cubic cm lead has 11.3g in compared to the 2.6 glass has. m/v= density. divide the masses over the volume and that will prove your answer.
8 0
3 years ago
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