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2 years ago

30 Points: How does it relate to one of these?

Mathematics

1 answer:

brilliants [131]2 years ago

8 0

Answer: the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Step-by-step explanation:

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How to solve interior angles of a triangle

harkovskaia [24]

Answer:

If you have one of the interior angles already there for you then the alternate interior angle would be the exact same

Step-by-step explanation: Alternate interior angles theorem

5 0

3 years ago

What basic trigonometric identity would you use to verify that sin^2x +cos^2x/cos x = sec x

gogolik [260]

<u>Answer:</u>

The basic identity used is $\bold{\sin ^{2} x+\cos ^{2} x=1}$ .

<u>Solution: </u>

In this problem some of the basic trigonometric identities are used to prove the given expression.

Let’s first take the LHS:

$\Rightarrow \frac{\sin ^{2} x+\cos ^{2} x}{\cos x}$

Step one:

The sum of squares of Sine and Cosine is 1 which is:

$\sin ^{2} x+\cos ^{2} x=1$

On substituting the above identity in the given expression, we get,

$\Rightarrow \frac{\sin ^{2} x+\cos ^{2} x}{\cos x}=\frac{1}{\cos x} \rightarrow(1)$

Step two:

The reciprocal of cosine is secant which is:

$\cos x=\frac{1}{\sec x}$

On substituting the above identity in equation (1), we get,

$\Rightarrow \frac{\sin ^{2} x+\cos ^{2} x}{\cos x}=\sec x$

Thus, RHS is obtained.

Using the identity $\sin ^{2} x+\cos ^{2} x=1$ , the given expression is verified.

6 0

3 years ago

To join a Yoga club there is a $100 annual fee and a $5 fee for each class you attend.

Leona [35]

Answer:

210

Step-by-step explanation: so 5 times 12 120 plus 100 is 210

5 0

3 years ago

Fine length of BC on the following photo.

MrMuchimi

Answer:

$BC=4\sqrt{5}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=AD^2+DC^2$

substitute the given values

$AC^2=16^2+8^2$

$AC^2=320$

$AC=\sqrt{320}\ units$

simplify

$AC=8\sqrt{5}\ units$

step 2

In the right triangle ACD

Find the cosine of angle CAD

$cos(\angle CAD)=\frac{AD}{AC}$

substitute the given values

$cos(\angle CAD)=\frac{16}{8\sqrt{5}}$

$cos(\angle CAD)=\frac{2}{\sqrt{5}}$ ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

$cos(\angle BAC)=\frac{AC}{AB}$

substitute the given values

$cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}$ ----> equation B

step 4

Find the value of x

In this problem

$\angle CAD=\angle BAC$ ----> is the same angle

equate equation A and equation B

$\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}$

solve for x

Multiply in cross

$(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units$

$DB=4\ units$

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

$BC^2=DC^2+DB^2$

substitute the given values

$BC^2=8^2+4^2$

$BC^2=80$

$BC=\sqrt{80}\ units$

simplify

$BC=4\sqrt{5}\ units$

7 0

3 years ago

a map shows the town where niko lives. the actual distance from niko's house to his school is 3miles, and measures one-half inch

garri49 [273]

6 miles = 1in

9 miles = 1 1/2in

12 miles = 2in

and it goes on, hope this helps!

6 0

3 years ago

Read 2 more answers