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labwork [276]
3 years ago
7

What is the measure of angle A' B' C'?​

Mathematics
1 answer:
Tresset [83]3 years ago
7 0

Answer:

40

Step-by-step explanation:

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Three students are sitting on a school bus. Sidney is 4 meters directly behind Quinn and 3 meters directly left of Ayana. Sidney
Ivenika [448]

Answer:

quinn,7 meters

Step-by-step explanation:

7 0
3 years ago
Anthony, Brian and Chris are hired for a job. If Anthony and Brian are working together they will finish the job in minutes. If
Tcecarenko [31]

Answer:

12.5

Step-by-step explanation:

Let  individual time taken by Anthony, Brian and Chris = t1 , t2 , t3 respectively

So, total time taken by A & B  = 1 / t1  + 1 / t2 , lets suppose = 10  {i}

Total time taken by A & C = 1 / t1  + 1 / t3 , lets suppose = 15 {ii}

Total time taken by B & C = 1 / t2  + 1 / t3 , lets suppose = 20 {iii}

From {i} , 1 / t2 = 10 - 1 / t1

From {ii} 1 / t1 = 15 - 1 / t3

By above two eqtns ,  1 / t2 = 10 - (15 - 1 / t3)

1 / t2 = 1 / t3 - 5

Putting in {iii} , 1 / t3 - 5 + 1 / t3 = 20

2 / t3 = 25

t3 = 12.5

5 0
3 years ago
I need to solve this using l'hopital's rule and logarithmic diferentiation.
arlik [135]

Yo sup??

For our convenience let h=x+1

therefore

when x tends to -1, h tends to 0

hence we can rewrite it as

\lim_{h \to \ 0 } (cos(h))^{(cot(h^2 )}

This inequality is of the form 1∞

We will now apply the formula

e^(^g^(^x^)^(^f^(^x^)^-^1^)^)

plugging in the values of g(x) and f(x)

e^{lim_{h \to \ 0}{(cot(h)^2(cos(h)-1))}

express coth² as cosh²/sinh² and also write cosh-1 as 2sin²(h/2)

(by applying the property that cos2x=1-sin²x)

After this multiply the numerator and denominator with h² so that we can apply the property that

\lim_{x \to \ 0 } sinx/x =1

Now your equation will look like this.

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2)*h^2)/(sin(h)^2*h^2)}

We will now apply the result

\lim_{x \to \ 0 } sinx/x =1

where x=h²

we get

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2)}

we now multiply the numerator and denominator with 4 so that we can say

\lim_{h^2 \to \ 0 } sin^2(h/2)/(h^2/4) = 1

e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2*4/4)}

=e^{lim_{h \to \ 0}{((cos(h)^2*2)/(4)}

Apply the limits and you will get

e^{cos(0)^2*2/4

=e^{1/2}

Hope this helps.

7 0
3 years ago
I need help and fast plz can someone help me
nikdorinn [45]

Answer:

Try D because a number times 10 to the power of anything is not that number!

Step-by-step explanation:


6 0
3 years ago
A county manager used a representative sample of two populations in the county to determine the number of pounds of trash thrown
REY [17]

Answer: He should have used the mean to determine the average because the data sets are approximately symmetric.

Step-by-step explanation

4 0
3 years ago
Read 2 more answers
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