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Vedmedyk [2.9K]
2 years ago
8

What is 1 * 2 = + * 1 = + + 2 2 + 1 8 5 / 9 Whoever answers it gets grevious

Mathematics
2 answers:
Helga [31]2 years ago
8 0

Answer:

Can you use a specific question please

Step-by-step explanation:

Vera_Pavlovna [14]2 years ago
7 0

Answer: = 17 7/

Step-by-step explanation:

8 1/5 + 9 1/2 = 177/

10

= 17 7/

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A recipe for 1 lb of trail mix calls for 2.5 oz of chocolate pieces. How many ounces of chocolate pieces are needed to make 5 lb
Murljashka [212]
5 lb of trail mix* (2.5 oz of chocolate pieces/ 1 lb of trail mix)= 12.5 oz of chocolate pieces.
(Note that the unit cancels out so you get the answer)

The final answer is 12.5 oz of chocolate pieces~
5 0
3 years ago
Simplify the expression. 24x + 32 + 4x + 3 What is the coefficient of x? What is the constant?
podryga [215]
28 is the coefficient of x and 35 is the constant
7 0
2 years ago
If it were two hours later,it would be half as long until midnight as it would be if it were an hour later. what time is it now.
mafiozo [28]

Answer:

its 10 am I think

Step-by-step explanation:

it says 2 hours until midnight. 12 - 2 = 10

7 0
3 years ago
Read 2 more answers
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
3 years ago
What’s the answer??????#9
zzz [600]

Answer:

A)

Step-by-step explanation:Quadratics

4 0
2 years ago
Read 2 more answers
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