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aleksklad [387]
2 years ago
7

Solve for xx/6 + ≥ −2​

Mathematics
1 answer:
Verizon [17]2 years ago
3 0
There is no answer because you didn’t put some thing after the + so it wouldn’t show nothing
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Pls help !!!!!!!!!!!!!!!!!!
Mashutka [201]
Maybe the answer is B
4 0
3 years ago
A golfer exerts a force of 15.0 N on a golf ball that has a mass of 0.42 kg. What is the golf
Gala2k [10]

Answer:

35.71m/s²

Step-by-step explanation:

using the formula force(F)=mass(m)×acceleration(a)

a =

\frac{f}{m}

F=15.0N

m=0.42kg

a=?

a=15÷0.42

a=35.71m/s²

5 0
3 years ago
Read 2 more answers
Can someone help please ? Answer correctly for brainliest and a thanks ! Don't answer if you don't know it please !
professor190 [17]
The first one would be approximately -0.8. It has a negative slope and the data points are fairly close together. The second one is almost a straight line so it would be very close to 1. I would say 0.97 The closer the data is to a straight line the closer the r value is to 1 or negative 1. Hope this helps.
6 0
3 years ago
What is the perimeter of the triangle?<br> O 22 units<br> O 30 units<br> O44 units<br> O 60 units
umka21 [38]

Answer:

22 units

Step-by-step explanation:

Add 12, 4 , and 6 and you will get 22

3 0
2 years ago
Read 2 more answers
(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
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