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uysha [10]
3 years ago
14

What is the value of x ?!

Mathematics
1 answer:
puteri [66]3 years ago
3 0

Answer:

4

Step-by-step explanation:

2x - 1 = 7

2x = 8

x = 4

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What is the quotient? Will give brainliest and will report absurd answers.
coldgirl [10]

Answer:

D

Step-by-step explanation:

When dividing fractions, multiply the first number by the reciprocal of the second number

2/5 ÷ 1/3

First, find the reciprocal of the second number: 1/3

To find the reciprocal, flip the numerator (top number) and denominator (bottom number)

1/3-->3/1

Now, multiply 2/5 and 3/1

2/5 * 3/1

Multiply across the numerators and denominators

6/5

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liberstina [14]

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4 0
3 years ago
Help me please this test is very important !
Lilit [14]

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d

Step-by-step explanation:

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6 0
3 years ago
Plz help me<br> 20 points
wel

Answer:

6

Step-by-step explanation:

Hope this helps! Pls give brainliest!

7 0
2 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
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