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3 years ago

Given: ΔABC; b= 10; c = 14, and ∠A = 54°. Find the length of side a to the nearest whole number.

1 answer:

algol133 years ago

8 0

We can solve for the length of side a to the nearest whole number using the Laws of Cosines such as the formula is shown below:
a²=b²+c²-2bcCosA
Solving for the value of a, we have:
a²=10²+14²-2(10)(14)cos54°
a²=131.42
a=11.46
The answer is 11.46 or 11.5.

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What slope is perpendicular to m= 5/8 m=-5/8 m=5/8 m=-8/5 m=8/5

MrRa [10]

Answer:

C) $m =\frac{-8}{5}$

The slope of the perpendicular line $m =\frac{-8}{5}$

Step-by-step explanation:

Explanation:-

Given that the slope of the given line is

$m =\frac{5}{8}$

The slope of the perpendicular line

= $\frac{-1}{m}$

The slope of the perpendicular line

= $\frac{-1}{\frac{5}{8} }$

Final answer

The slope of the perpendicular line

$m =\frac{-8}{5}$

7 0

3 years ago

Plz answer. only if correct

earnstyle [38]

Answer:10001.5%+10303990=f(g+2)

Step-by-step explanation:

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3 years ago

Th sum of three numbers is 64. The second number is 3 more than the first. The third number is

ladessa [460]

Three numbers can be defined as x, y, and z.
x + y + z = 64
y = x+3
z = 2x - 11
.
substitute for y and z
x + (x+3) + (2x-11) = 64
4x -8 = 64
4x = 72

x = 18
.
y = x+3 = 21
z = 2x -11 = 2(18) -11 = 36-11 = 25
.
x + y + z = 18 + 21 + 25 = 64
Correct.
.
Answer: The three numbers are 18, 21, and 25.

8 0

3 years ago

Find the probability of the following events , when a dice is thrown once:

Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

7 0

3 years ago

Read 2 more answers

Consider a uniform distribution from aequals4 to bequals29. (a) Find the probability that x lies between 7 and 27. (b) Find th