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3 years ago

Chas build a circular pattern for his dog The radius of the pen is 9 feet how much fencing did Chaz use for the pen

Mathematics

1 answer:

kicyunya [14]3 years ago

3 0

Answer:

The fencing Chaz used for the pen is 56.67 feet

Step-by-step explanation:

Fencing interprets to circumference

If the radius of the pen is 9 ft

circumference = $2\pi r$

= 2*(22/7)*9

= 396/7

= 56.67 ft

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Which is equivalent to 437 O A 7 B. 12 OC 16 OD. 64 E. 81

soldier1979 [14.2K]

D 64 if this is right please give me brainlist

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3 years ago

You are visiting a rainforest, but unfortunately,your insect repellent has run out. As a result, at each second, a mosquito land

zmey [24]

Answer:

Step-by-step explanation:

Prob at each second, a mosquito lands on your neck =0.5

Prob it bites/it lands = 0.2

Prob it does not bite/it lands = 0.8

Hence in one second prob for getting a bite = $0.5(0.2) = 0.01$

i.e. 1/100 chance of getting bitten in one second

In 100 seconds number of bites $=1$

Expected time between successive bites = 100

6 0

4 years ago

Simplify: 18q2−45q+25/9q2−25.

Svet_ta [14]

Answer:

$\frac{(6q-5)}{(3q+5)}$

Step-by-step explanation:

Given: $\frac{18q^{2}-45q+25 }{9q^{2}-25 }$

Factorizing the numerator and the denominator, we have:

18 $q^{2}$ - 45q + 25 = 18 $q^{2}$ - 30q - 15q + 25

= (3q - 5)(6q - 5)

and

9 $q^{2}$ - 25 = (3q - 5)(3q + 5)

So that,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(3q-5)(6q-5)}{(3q-5)(3q+5)}$

= $\frac{(6q-5)}{(3q+5)}$

Therefore,

$\frac{18q^{2}-45q+25 }{9q^{2}-25 }$ = $\frac{(6q-5)}{(3q+5)}$

5 0

3 years ago

99 POINTS, NEED HELP WITH ALGEBRA

Tanya [424]

1 is in picture above.

$average \: rate = \frac{f(4) - f(1)}{4 - ( - 1)} = \frac{5}{4 - ( - 1)} = 1$

3. In the picture...

4. In the picture...

5. Parts A and B are true.

$average \: rate = \frac{f(0) - f( - 10)}{0 - ( - 10)} = \frac{( - 6) - 184}{0 - ( - 10)} = \frac{ - 190}{10} = - 19$

7. The axis symmetry equation of

$y = a {x}^{2} + bx + c$

$x = \frac{ - b}{2a}$

So it would be x=-3/4

$y = {(x - 6)}^{2} + 10$

T=(6, 10) vertex coordinates...

4 0

3 years ago

Read 2 more answers

The percent of working students increased 8.1 to 40.5 what was the present prior to increase

Dmitrij [34]

Answer:

32.4

Step-by-step explanation:

prior + 8.1 = 40.5 . . . . . . seems to model the problem statement

prior = 32.4 . . . . . . . subtract 8.1 from both sides

Prior to the increase the percent was 32.4.

_____

Comment on the problem statement

When you're talking about a percentage increase in a percentage, it is almost never clear whether you're talking about the percentage of the underlying number, or the percentage of the percentage.

Here, we assume the 8.1 is a percentage of working students, not a percentage of the percentage of workings students. If you actually intend the latter, the percentage before the increase was about 37.465%.

6 0

3 years ago