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3 years ago

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Mr. Tuyen uses 5/8 of a tank of gas each week to drive to and from his job. How many tanks of gas does Mr. Tuyen use in 5 weeks?

Write your answer in two different ways

1 answer:

Leokris [45]3 years ago

3 0

Mr.Tuyen uses 3 1/8 tanks of gas in 5 weeks. The improper fraction version would be 25/8 if that is what you're looking for

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Write the next three terms of the arithmetic sequence <br> 1) -7, -14, -21, -28

Sunny_sXe [5.5K]

Step-by-step explanation:

This sequence is the table of 7 it is negative we will continue so

-7 -14 -21 -28 -35 -42 -49

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3 years ago

Annie received a 10% raise last year. If her salary now is $60,000, what was her salary last year?

sergeinik [125]

$6,000 was her salary last year:))

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3 years ago

Solve inequalities.

grigory [225]

Answer:

10. 7n - 1 < -8

Isolate the variable, n. Do the opposite of PEMDAS. Treat the < as equal sign, what you do to one side, you do to the other. First, add 1 to both sides:

7n - 1 (+1) < - 8 (+1)

7n < - 8 + 1

7n < - 7

Isolate the variable, n. Divide 7 from both sides:

(7n)/7 < (-7)/7

n < -7/7

n < -1

n < -1 is your answer.

11. 3 > -7v + 4v

Combine like terms, then isolate the variable, v. First, add -7v and 4v together.

3 > (-7v + 4v)

3 > (4v - 7v)

3 > (-3v)

Isolate the variable, v. Divide -3 from both sides. Note that since you are dividing a negative number, you must flip the sign:

(3)/-3 > (-3v)/-3

3/-3 > v

-1 < v

v > -1 is your answer.

~

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3 years ago

Rewrite in simplest terms: 4(-5u +3) – 4(8u - 1)

viktelen [127]

Answer:

You distribute then combine

your answer is <u><em>-52u+16</em></u>

Step-by-step explanation:

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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