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vekshin1
3 years ago
10

Write an inequality for ten less than a number is at least 5

Mathematics
1 answer:
alexandr1967 [171]3 years ago
5 0

Step-by-step explanation:

Let the number be x

so,

10 - x ≥ 5

Hope it helps ya

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Please help. I don’t understand what to do
11111nata11111 [884]

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=23.9\\ h=100 \end{cases}\implies V=\cfrac{\pi (23.9)^2(100)}{3} \\\\\\ V=\cfrac{57121\pi }{3}\implies V\approx 59816.97\implies \stackrel{\textit{rounded up}}{V=59817} \\\\[-0.35em] ~\dotfill

now, for the second one, we know the diameter is 10, thus its radius is half that or 5.

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=5\\ V=225 \end{cases}\implies 225=\cfrac{\pi (5)^2 h}{3}\implies 225=\cfrac{25\pi h}{3} \\\\\\ \cfrac{225}{25\pi }=\cfrac{h}{3}\implies \cfrac{9}{\pi }=\cfrac{h}{3}\implies \cfrac{27}{\pi }=h\implies 8.59\approx h\implies \stackrel{\textit{rounded up}}{8.6=h}

6 0
4 years ago
9р - Зр equivalent expression​
Vesnalui [34]

Answer:

6p

Step-by-step explanation:

9p - 3p needs to have the like terms combined. 9 - 3 is 6, and since you have to keep the p, the answer is 6p.

7 0
3 years ago
When completely factored, x^2 + x - 6 is equivalent to which of the following? A. (x + 1)(x – 6) B. (x + 3)(x – 2) C. (x + 6)(x
Mariana [72]
X^2+x-6

x^2-2x+3x-6

x(x-2)+3(x-2)

(x+3)(x-2)

5 0
4 years ago
Provide an expression that is equivalent to 5(3-2) ?
VMariaS [17]

Answer:

5

Step-by-step explanation:

distribute the negative

5*3 + 5(-2)

15-10

5

3 0
3 years ago
What is the correct solution set for the following graph?
Molodets [167]

Answer:

The correct solution set is a point in Quadrant II.

Step-by-step explanation:

You can see that if you extend the lines, they will eventually intersect at only one point. The quadrants start with I in the upper right corner, go II in the upper left corner, III in the bottom left corner, and IV in the bottom right corner. You can see that they will eventually intersect in the upper left corner, Quadrant II.

7 0
4 years ago
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