All categories

2 years ago

Find the measure of angle A

Mathematics

2 answers:

Digiron [165]2 years ago

6 0

Answer:

The correct answer option is b. m∠B = 61°.

Step-by-step explanation:

We are given a right angles triangle with an angle ABC equal to 29 degrees and side length BC known to be equal to 15.

We are supposed to find the measure of angle A.

We know that the sum of all three angles of a triangle must be 190 degrees. Since it is a right angled triangle, so ∠C = 90° and ∠B = 29°.

A° + B° + C° = 180°

A° + 29° + 90° = 180°

A° = 180° - 119°

A° = 61°

Salsk061 [2.6K]2 years ago

4 0

the angle c is 90°, it is a right triangle. (you see from the picture).

the sum of the interior angles are 180° (in all triangles)

180 - 90 - 29 =

61° (angle A)

----------------------

90 + 61 + 29 = 180°

You might be interested in

Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$

$C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35$

$C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21$

$C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1$

For all the calculations just recall that 4! =24 and 5!=120.

6 0

2 years ago

How much of a radioactive kind of ruthenium will be left after 8 hours if you start with 3,888

ziro4ka [17]

Answer:

Step-by-step explanation:

As an example, iodine-131 is a radioisotope with a half-life of 8 days. It decays by beta particle emission into xenon-131. After eight days have passed, half of the atoms of any sample of iodine-131 will have decayed, and the sample will now be 50% iodine-131 and 50% xenon-131.

50 grams to 25 grams is one half-life. 25 grams to 12.5 grams is another half-life. So, for 50 grams to decay to 12.5 grams, two half-lives, which would take 36 days total, would need to pass. This means each half-life for element X is 18 days.

7 0

2 years ago

Read 2 more answers

Need help combining like terms

yaroslaw [1]

Answer:

-10x+7.5

Step-by-step explanation:

-2.5(4x-3)

Multiply each term by -2.5

-2.5 * 4x = -10x

-2.5 * 3 = 7.5