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Harman [31]
2 years ago
11

Find the area of triangle ABC. Give your answer correct to 1 decimal place AC=7 Cb=12 a=72 degrees b=59 degrees

Mathematics
1 answer:
Tems11 [23]2 years ago
3 0

Answer:

31.7 cm²

Step-by-step explanation:

The area (A) of the triangle is calculated as

A = \frac{1}{2} absinC

where a, b are the 2 sides given and C the angle between them

∠ C = 180° - (72 + 59)° = 180° - 131° = 49° , thus

A = 0.5 × 12 × 7 × sin49° ≈ 31.7 cm² ( to 1 dec. place )

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steposvetlana [31]

Answer:

Hi there!

The correct answer to this question is: x=3

Step-by-step explanation:

because these two triangles are similar, that means the hypotenuse are similar. which means you can set the sides equal to each other like this:

13 = 4x + 1   minus one on both sides and you should get 4x = 12 and then divide 4 on both sides and you should finally get x = 3

7 0
3 years ago
What is the solution of the inequality shown below X-3>7
True [87]

Step-by-step explanation:

x-3>7 =>

x>10

hope this helps

5 0
2 years ago
Answer all for Brainliest (easy math) 5 questions for 30 points
sweet-ann [11.9K]

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2) Substitute the known values into the equation. There are 7 people at $8.25 a person, which is a total of $57.75 <u>I used 7 because he invites 6 friends, plus himself. </u>He only has $50 on his gift card, so he won't have enough to cover the cost. The answer is A.

3) They start out with 40lbs and recieve 120 more. They have 160lbs in total. If they sell 8lbs a day for 5 days, thats 8*5 or 40lbs total. 160 - 40 = 120lbs left after 5 days. I'd say the answers are A, B, and D.

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I hope this helps you! I'm sorry if I got any of them wrong!

7 0
3 years ago
Which sum or difference identity would you use to verify that cos (180° - q) = -cos q?
Phantasy [73]

Answer:

\cos (a-b)=\cos a \cos b+\sin a \sin b

Step-by-step explanation:

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We have to write which identity we will use to prove the given statement.

Consider \cos (180^{\circ}-q)=-\cos q

Take left hand side of given expression \cos (180^{\circ}-q)

We know

\cos (a-b)=\cos a \cos b+\sin a \sin b

Comparing , we get, a= 180° and b = q

Substitute , we get,

\cos (180^{\circ}-q)=\cos 180^{\circ}  \cos (q)+\sin q \sin 180^{\circ}

Also, we know \sin 180^{\circ}=0 and \cos 180^{\circ}=-1

Substitute, we get,

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\cos (180^{\circ}-q)=-\cos (q)

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7 0
3 years ago
Read 2 more answers
Determine the values of x on which the function f(x)=2x^2-x-15/4x^2-12x is discontinuous and verify the type of discontinuity at
melomori [17]
<span>Determine the values of x on which the function f(x)=2x^2-x-15/4x^2-12x is discontinuous and verify the type of discontinuity at each point.

A.There is a vertical asymptote at 0 and a hole at 3.
B.There are vertical asymptotes at -5/2 & 0 and a hole at 3.
C.There is a vertical asymptote at 3 and a hole at 0.
D.There are vertical asymptotes at 0 & 3.
</span>
Result:
(-7/4)x^2-13x

The roots are:
x= -52/7
x=0

See attached pictures.

6 0
3 years ago
Read 2 more answers
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