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3 years ago

9

Derek wants to find the area of this triangle. Which unit of measurement should he use? A. yd2 B. yd C. yd3

1 answer:

jek_recluse [69]3 years ago

3 0

It is c cause of the right angle

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Solve the equation.b²=45

Stolb23 [73]

Answer:

b² ≈ 6.71

Step-by-step explanation:

The $\sqrt{45}$ = 6.708

6.708 ⇒ 6.71

and $6.71^{2}$ ≈ 45.0241

b² = 6.71

5 0

3 years ago

Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1

IceJOKER [234]

Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

<u>Step-by-step explanation:</u>

Here we have , $cosxtanx - 2 cos^2 x=-1$ . Let's solve :

⇒ $cosxtanx - 2 cos^2 x=-1$

⇒ $cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1$

⇒ $sinx = 2 cos^2 x-1$

⇒ $sinx = 2 (1-sin^2x)-1$

⇒ $sinx = 1-2sin^2x$

⇒ $2sin^2x+sinx-1=0$

By quadratic formula :

⇒ $sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}$

⇒ $sinx = \frac{-1 \pm3}{4}$

⇒ $sinx = \frac{1}{2} , sinx =-1$

⇒ $sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}$

⇒ $x=\frac{\pi}{6} , x=\frac{3\pi}{2}$

But at $x=\frac{3\pi}{2}$ we have equation undefined as $cos\frac{3\pi}{2}=0$ . Hence only solution is :

⇒ $x=\frac{\pi}{6}$

Since , $sin(\pi -x)=sinx$

⇒ $x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}$

Now , General Solution is given by :

⇒ $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$

Therefore , Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

3 0

3 years ago

I don't understand it maybe y'all do

kap26 [50]

Right angle triangle

4 0

3 years ago

Read 2 more answers

Any of you guys know this answer?

Fantom [35]

The answer would be 10

5 0

3 years ago

What is the greatest possible error if bruce measured a buckle as 3.2 cm using the ruler? what is the greatest possible error? 0

solniwko [45]

Answer:

I believe it's 5.0

Step-by-step explanation:

I believe so if its not I'm sorry

5 0

1 year ago