All categories

3 years ago

If A=78º, b=16, c=7, what is the area of this triangle?

Mathematics

1 answer:

Andrei [34K]3 years ago

8 0

Answer:

= 54.78 square units

Step-by-step explanation:

We can calculate the area of the triangle using the formula;

Area = 1/2abSinθ ; where a and b is the sides making the angle θ.

Therefore;

Area = 1/2 bc sin C

= 1/2 × 16 × 7 × Sin 78

= 8 × 7 × sin 78

<u>= 54.78 square units</u>

You might be interested in

A college basketball coach has 12 players on his team. eight players are receiving scholarships, and four are not. the coach dec

EleoNora [17]

Solution: The given problem is a binomial distribution.

The probability that a team member receives a scholarship is $\frac{8}{12} = 0.67$

Therefore, the given problem follows binomial with n = 5 and p = 0.6667

Now the probability that 4 of 5 players selected are on scholarship is:

$P(x=4) = \binom{5}{4} \times 0.6667^{4} (1-0.6667)^{5-4}$

$=5 \times 0.1976 \times 0.3333$

$=0.329$

Therefore, the probability that 4 of 5 players selected are on scholarship is 0.329

5 0

3 years ago

0.5+0.3-(-0.5--0.3) I AM SOOOO SRRY GUYS

11Alexandr11 [23.1K]

Answer:

0.5+0.3-(-0.5-0.3)

=1.6

3 0

2 years ago

What is the correct solution for 8x = 56

Mars2501 [29]

Answer:

Step-by-step explanation:

8(7)=56

4 0

3 years ago

Read 2 more answers

A jeweler wants to make 14 grams of an alloy that is precisely 75% gold.. The jeweler has alloys that are 25% gold, 50% gold, &a

Goryan [66]

Given that the jeweler has alloys that are 25% gold, 50% gold, and 82% gold.

As he wants to make 14 grams of an alloy by adding two different alloys that is precisely 75% gold, so one alloy must have a percentage of gold more than 75%.

One alloy is 82% gold and, the second can be chosen between 25% gold, 50% gold, so there are two cases.

Case 1: 82% gold + 50% gold

Let x grams of 82% gold and y grams of 50% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 50% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times y \\\\$

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 50/100 \times (14-x)$ [as x+y=14]

$\Rightarrow 75 \times 14 = 82 \times x + 50 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 50 \times14-50\times x \\\\\Rightarrow 75 \times 14 = 32 \times x + 50 \times14 \\\\\Rightarrow 32 \times x =75 \times 14 - 50 \times14 \\\\$

$\Rightarrow x =(25 \times 14)/32=10.9375$ grams

and $y = 14-x= 14-10.9375=3.0625$ grams.

Hence, 10.9375 grams of 82% gold and 3.0625 grams of 50% gold added to make 14 grams of 75% gold.

Case 2: 82% gold + 25% gold

Let x grams of 82% gold and y grams of 25% gold added to make x+y=14 grams of 75% gold, so

75% of 14 = 82% of x + 25% of y

$\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times y \\\\\Rightarrow 75/100 \times 14 = 82/100 \times x + 25/100 \times (14-x) \\\\ \Rightarrow 75 \times 14 = 82 \times x + 25 \times (14-x) \\\\\Rightarrow 75 \times 14 = 82 \times x + 25 \times14-25\times x \\\\\Rightarrow 75 \times 14 = 57 \times x + 25 \times14 \\\\\Rightarrow 57 \times x =75 \times 14 - 25 \times14 \\\\$