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2 years ago

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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If the triangles are congruent, then X=?

steposvetlana [31]

Answer:

Hi there!

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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

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2 years ago

What is the solution of the inequality shown below X-3>7

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Step-by-step explanation:

x-3>7 =>

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2 years ago

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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:

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

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

Hence, use difference identity to prove the given result.

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