The perimeter of pqrs is 84. Find the length of each side qr=3(rs)

How do I solve this question 6x-5x=0

Tpy6a [65]

Answer:

x=0

Step-by-step explanation:

0-0=0

4 0

3 years ago

Read 2 more answers

Plz help me I need tu turn this is by midnight!!

expeople1 [14]

Answer:

The last answer.

Step-by-step explanation:

It looks to be the correct answer because for the one above you can't simplify 27 by 5.

4 0

3 years ago

Read 2 more answers

Solve for x.<br><br> -98=4x-50

alexdok [17]

The answer to your question is x= -12 hope this helps ! Makes sure you put the negative !

3 0

2 years ago

MARKING BRANLY You want to plant a flower garden in your yard so that you can make a beautiful bouquet to put on the alter at ch

BabaBlast [244]

Answer:

1. 3 packs of daffodils and 4 packs of tulips

2. then you would have 24 of each

3. $2.29

4. $37.54

5. $34.61

6.$2.93 was saved

7. 3 tulips

8. 12 daffodils

9. 9 tulips

10. 12 daffodils and 15 tulips

theres 2 left. ill let someone else do it

Step-by-step explanation:

3. 4.50*4=18 5.75*3=17.25

18+17.25 -> 35.25*.065=2.29 rounded

4. 35.25+2.29=37.54

5. 35.25-2.75->32.50*.065=2.11 2.11+32.50=34.61

6. 37.54-34.61=2.93

7. 6:8 -> 3:4 ratio -> 3:4

8. 8+4=12

9. 6+3=9

10. 24-12=12 24-9=15

6 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

3 years ago