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Andreyy89
3 years ago
10

Approximately how many square inches of space are blank?

Mathematics
1 answer:
ddd [48]3 years ago
3 0

Answer:

blank area = 1.5π units²

Step-by-step explanation:

area of full circle = πr² = π6² = 36π units²

blank area = (15/360)(36π) = 1.5π units²

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RideAnS [48]

Answer:

yes at negative x-coordnintn

Step-by-step explanation:

8 0
2 years ago
BD bisects /ABC. Find the value of x.​
Anvisha [2.4K]

Answer:

x=8

Step-by-step explanation:

we know that

If BD bisects angle ABC

then

m∠ABD=m∠DBC

substitute the given values

(2x+7)\°=(4x-9)\°

Solve for x

4x-2x=7+9

2x=16

x=8

7 0
3 years ago
Please help!! due by 1:30!
dem82 [27]
An = 4n + 4
AKA C is the correct answer.

say n is the number of the month after the start of the year

January would be a1 = 4*1 + 4 = 8
February would be a2 = 4*2 + 4 = 12 

and so on
5 0
4 years ago
Paul received a 12,000 loan from the bank. The bank charges his 6.99% yearly interest rate. 4 years, how much money does Paul ow
andre [41]
Principal amount = 12,000
Annual interest rate (r) = 6.99% = 0.0699
Time (years) = 4 years
Number of installments (t) = 12*4 = 48 months

Monthly payment, A = P/D

Where,
D= {(1+r/12)^t-1}/{r/12*(1+r/12)^t} = {(1+0.0699/12)^48-1}/{0.0699/12(1+0.0699/12)^48} = 42.47

Therefore, A = 12000/42.47 = 282.59

Total payments after 4 years = 282.59*4*12 = 13,564.17

Interest owed = 13,564.17 - 12,000 = 1,564.17
5 0
3 years ago
Read 2 more answers
Calculate the limit of the function with L'Hospital rule​
mr_godi [17]

Answer:

L=24

Step-by-step explanation:

L'Hopital's Rule for Evaluating Limits:

Rule is that if \lim_{x \to a} \frac{f(x)}{g(x)} takes \frac{0}{0} or \frac{\infty}{\infty} form, then,

\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}

where f'(x)=\frac{df(x)}{dx} and g'(x)=\frac{dg(x)}{dx}

Now coming to the problem,

L= \lim_{x \to \frac{\pi}{6} } \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3} )}

Here f(x)=cot^{3}x-3cotx and g(x)=cos(x+\frac{\pi}{3} )

Substituting x=\frac{\pi}{6} in f(x) and g(x),

f(\frac{\pi}{6})=cot^{3}\frac{\pi}{6}-3cot\frac{\pi}{6}\\=3\sqrt{3}-3\sqrt{3}\\ =0

g(\frac{\pi}{6})=cos(\frac{\pi}{6}+\frac{\pi}{3})\\=cos\frac{\pi}{2}\\=0

Since L takes the form \frac{0}{0}, using l'hopital's rule

L= \lim_{x \to \frac{\pi}{6}} \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3})}= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}x(-cosec^{2}x)-3(-cosec^{2}x)}{-sin(x+\frac{\pi}{3})}

now substituting x=\frac{\pi}{6} ,

L= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}\frac{\pi}{6}(-cosec^{2}\frac{\pi}{6})-3(-cosec^{2}\frac{\pi}{6})}{-sin(\frac{\pi}{6}+\frac{\pi}{3})}\\=\frac{3*3^{2}(-2^{2})+3(2^{2})}{-1}\\=24

6 0
3 years ago
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