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4 years ago

6

A new is being built in . The perimeter of the rectangular playing field is yards. The length of the field is yards less than th

e width. What are the dimensions of the playing field?

1 answer:

Dennis_Churaev [7]4 years ago

3 0

Answer:

1-------------12

Step-by-step explanation:

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Compute the square root of√144

zloy xaker [14]

The square root of 144 is 12

4 0

3 years ago

Read 2 more answers

(b) Kevin ran 4 miles more than Steve ran. The sum of their distances is 26 miles. How far did Steve run? The domain of the solu

kotegsom [21]

S=Steve. S+4=Kevin. S+s+4=26. Combine: 2s+4=26. Subtract 4: 2s=22. Divide by 2: s=11. Steve ran 11 miles. :)

5 0

3 years ago

WILL MARK AS BRAINLIEST IF CORRECT!!!

gizmo_the_mogwai [7]

Answer:

C

Step-by-step explanation:

If I'm not mistaken, she wants to ride atleast 85 miles per week. If you add up all of the values from going to and from the school, you would get 62.5 miles. So for it to add up to 85 miles, she would have to atleast do 9 trips around the park.

These numbers I believe do not matter however, all you need to do is put in C for greater than or equal to 85.

4 0

3 years ago

Read 2 more answers

At the start of the year, 15 chameleons were introduced into a zoo. The population of chameleons is expected to grow at a rate o

bearhunter [10]

Answer:

option-B

Step-by-step explanation:

We are given

At the start of the year, 15 chameleons were introduced into a zoo

so, $P_0=15$

The population of chameleons is expected to grow at a rate of 41.42% every year

so, r=0.4142

and x represents the number of years since the chameleons were introduced into the zoo

now, we can set equation to find total population

and we get

$P(x)=P_0(1+r)^x$

now, we can plug values

$P(x)=15(1+0.4142)^x$

$P(x)=15(1.4142)^x$

Average rate of change between 2 years and 4 years:

we can use formula

$A_1=\frac{P(4)-P(2)}{4-2}$

now, we can plug values

$A_1=\frac{15(1.4142)^{4}-15(1.4142)^{2}}{4-2}$

$A_1=14.99914$

Average rate of change between 4 years and 6 years:

we can use formula

$A_2=\frac{P(6)-P(4)}{6-4}$

now, we can plug values

$A_2=\frac{15(1.4142)^{6}-15(1.4142)^{4}}{6-4}$

$A_2=29.99770$

Average rate of change between 6 years and 8 years:

we can use formula

$A_3=\frac{P(8)-P(6)}{8-6}$

now, we can plug values

$A_3=\frac{15(1.4142)^{8}-15(1.4142)^{6}}{8-6}$

$A_3=59.99425$

now, we will check each options

option-A:

we can see that

$A_3-A_2=30$

$A_3-A_2=30$

So, this is FALSE

option-B:

$A_1=\frac{1}{2}A_2$

So, this is TRUE

option-C:

This is FALSE

option-D:

we got

$A_1=\frac{1}{2}A_2$

so, this is FALSE

5 0

3 years ago

Help me please due in 2 mins!!!!,!?????<br> Which of these is a geometric sequence??

Sloan [31]

Answer:

option d is correct

Step-by-step explanation:

hope this helps ;)

6 0

2 years ago