All categories

3 years ago

What is the value of x ?!

Mathematics

1 answer:

puteri [66]3 years ago

3 0

Answer:

Step-by-step explanation:

2x - 1 = 7

2x = 8

x = 4

You might be interested in

What is the quotient? Will give brainliest and will report absurd answers.

coldgirl [10]

Answer:

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

5 0

3 years ago

Are the triangles congruent, if so by what theorem/postulate

liberstina [14]

Yes, SSS

PR are already a given side. PS and PQ are also given sides. Lastly, RS and RQ are given sides.

4 0

3 years ago

Help me please this test is very important !

Lilit [14]

Answer:

Step-by-step explanation:

5>2n+11

5-11>2n

-6>2n

-6/2>n

n<or equal to -3

bubble is filled pointing left

6 0

3 years ago

Plz help me<br> 20 points

wel

Answer:

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