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Ostrovityanka [42]
3 years ago
14

Which algebraic expression has a term with a coefficient of 3?

Mathematics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

D. 3y+1

Step-by-step explanation:

The coefficient is the number in front of the variable

3y has a coefficient of 3

3 ( y-6) has a factor of 3 since it is multiplied by the other term ( y-6)

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Somebody please help​
iren [92.7K]

Answer:

Yes, No, No, No

Step-by-step explanation:

To decide whether the point lies on the circle, what you need to do it simply substituting the x and y values into the equation and check if it add up to be 25

(-5)² + 0² = 25, = RHS  [Yes]

1² + (√7)² = 8, ≠ RHS [No]

(√21)² + (-3)² = 30, ≠RHS [No]

0² + 7² = 49, ≠RHS [No]

5 0
3 years ago
PLEASE HELP I NEED THE ANSWER ASAP
Andreas93 [3]

Answer:

x>3, the last option with the open circle

Step-by-step explanation:

7 0
3 years ago
Simon (Mr. Lovett) completed 4 sets of chin-ups. Every set he reached a different prime number of chin-ups. Knowing this, what n
Vsevolod [243]

Answer:

answer is b!

Step-by-step explanation:

hope this helps!!!

8 0
3 years ago
If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :    

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

(\text{Average velocity})_{0.1\ s}=696\ ft/s

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

5 0
4 years ago
Please solve. v/7 = 8
Naya [18.7K]

Answer:

V=56

Step-by-step explanation:

times both sides by 7

3 0
3 years ago
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