Anyone know the answer?

In triangle ABC find b to the nearest degree given a=7 b=10 and c is a right triangle

eimsori [14]

Answer:

∠B = $55.00797173^{\circ}$

Step-by-step explanation:

Given : a=7 b=10 and C is a right triangle

solution :

refer the attached figure .

Since ΔACB is right angled triangle so we will use trigonometric ratios .

$tan\theta=\frac{perpendicular}{base}$

Perpendicular = b =10

Base=a=7

Putting value in formula:

$tanB=\frac{10}{7}$

$tanB=1.42857$

$B=tan^{-1}1.42857$

$B=55.00797173^{\circ}$

Thus ∠B = $55.00797173^{\circ}$

4 0

3 years ago

Read 2 more answers

Can someone rewrite 6x + 2y = -144 as a slope intercept equation?

SVEN [57.7K]

Answer:

y = -3x -72

Step-by-step explanation:

$6x+2y=-144\\\\2y=-6x-144\\\\y=\frac{-6}{2}x-\frac{144}{2}\\\\y=-3x-72$

4 0

3 years ago

You always need some time to get up after the alarm has rung. You get up from 10 to 20 minutes later, with any time in that inte

Mama L [17]

Answer:

a) P(x<5)=0.

b) E(X)=15.

c) P(8<x<13)=0.3.

d) P=0.216.

e) P=1.

Step-by-step explanation:

We have the function:

$f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \, otherwise }} \right.$

a) We calculate the probability that you need less than 5 minutes to get up:

$P(x$

Therefore, the probability is P(x<5)=0.

b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.

E(X)=15.

c) We calculate the probability that you will need between 8 and 13 minutes:

$P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3$

Therefore, the probability is P(8<x<13)=0.3.

d) We calculate the probability that you will be late to each of the 9:30am classes next week:

$P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6$