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3 years ago

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In how many ways may one A, three B’s, two C’s, and one F be distributed among seven

1 answer:

astraxan [27]3 years ago

5 0

7 different ways is the answer

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Solve the triangle. Round your answers to the nearest tenth. Law of sines 6 A. M∠C=34 m ∠ C = 34 , b=25 b = 25 , c=16 c = 16 B.

PolarNik [594]

Answer:

$A = 43.0^o$

$B = 55.0^o$

$BC = 20.0$

Step-by-step explanation:

I will solve this question with the attached triangle

From the attachment, we have:

$\angle C =82^o$

$AB = 29$

$AC = 24$

Required

Solve the triangle

First, we calculate $\angle B$ using sine law:

$\frac{AC}{\sin(B)} = \frac{AB}{\sin(C)}$

This gives:

$\frac{24}{\sin(B)} = \frac{29}{\sin(82^o)}$

$\frac{24}{\sin(B)} = \frac{29}{0.9903}$

Cross multiply

$\sin(B) * 29 = 24 * 0.9903$

$\sin(B) * 29 = 23.7672$

Divide both sides by 29

$\sin(B) = 0.8196$

Take arcsin of both sides

$B = \sin^{-1}(0.8196)$

$B = 55.0^o$

Next, calculate $\angle A$ using:

$A + B + C = 180^o$ --- angles in a triangle

$A + 55^o + 82^o = 180^o$

Collect like terms

$A = 180^o-55.0^o - 82^o$

$A = 43.0^o$

Next, calculate BC using sine laws

$\frac{BC}{\sin(A)} = \frac{AB}{\sin(C)}$

This gives:

$\frac{BC}{\sin(43.0)} = \frac{29}{\sin(82)}$

$\frac{BC}{0.6820} = \frac{29}{0.9903}$

Make BC the subject

$BC = \frac{29*0.6820}{0.9903}$

$BC = 20.0$

6 0

3 years ago

One number is 3 less than another number if their sum is 49

Musya8 [376]

One number is 26 the other is 23

x+y=49

Y=x+3,X (now substitute it in the equation)

x+3+x=49

2x+3=49

2x=46. X=23

Y=x+3. Y=23+3 =26 for checking 23+26=49

6 0

2 years ago

Both the football and volleyball teams have games today. The football team plays every 7 days. The volleyball team plays every 3

Allisa [31]

They will have games on the same day in 21 days

6 0

3 years ago

Simplify: (3x2 - 4x + 6) + (7x - 2)

V125BC [204]

Answer:

11x-2

Step-by-step explanation:

(3x2 - 4x + 6) + (7x - 2) This is the breakdown. (6 4x + 6)= 4x. + (7x-2) = 11x-2

8 0

3 years ago

Helppppppp pleasssseeeeee

Svet_ta [14]

$14mi-5\frac{7}{10}mi=[(14-5)-\frac{7}{10}]mi=(9-\frac{7}{10})mi=(8+1-\frac{7}{10})mi\\\\=(8+\frac{10}{10}-\frac{7}{10})mi=(8+\frac{10-7}{10})mi=(8+\frac{3}{10})mi=\boxed{8\frac{3}{10}mi}\\\\other\ method\\\\14mi-5\frac{7}{10}mi=14mi-5.7mi=(14-5-0.7)mi\\\\=(9-0.7)mi=\boxed{8.3mi}$

6 0

3 years ago