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Marizza181 [45]
2 years ago
6

Identify this problem plssss

Mathematics
1 answer:
JulsSmile [24]2 years ago
8 0

Answer:

The answer is cot(x)

Step-by-step explanation:

<h3>Greetings !</h3>

\tan(x)  \csc {}^{2} (x)  -  \tan(x)

Rewrite using trig identity

-  \tan(x)  +  \csc {}^{2} (x) \tan(x)

use the Pythagorean idea

\csc {}^{2} (x)  = 1 +  \cot {}^{2} (x)

-  \tan(x)  + (1 +  \cot {}^{2} (x)) \tan(x)

simplify

\cot {}^{2} (x)   \tan(x)

use basic trigonometric identity

\tan(x    )  =  \frac{1}{ \cot(x) }

simplify

\cot {}^{2} (x)  \frac{1}{ \cot(x) }

gives you

\cot(x)

Hope it helps!

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What is x + x in the mathematics
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Answer:

X+x =2x

Step-by-step explanation:

I think this ans may help you

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Find the average rate of change of f(x)=2x^2-7x from x=2 to x=6
scoundrel [369]

Answer:

Step-by-step explanation:

The average rate of change of a function f between a and b (a< b) is :

R =[f(b)-f(a)] ÷ (a-b)

●●●●●●●●●●●●●●●●●●●●●●●●

Let R be the average rate of change of this function

f(6) = 2×6^2 - 7×6 = 72-42 = 30

f(2) = 2×2^2 - 7×2 = 8-14 = -6

R = [f(6) - f(2)]÷ (6-2)

R = [30-(-6)] ÷ 4

R = -36/4

R = -9

■■■■■■■■■■■■■■■■■■■■■■■■■

The average rate of change of this function is -9

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Write the fraction 7/33 as a decimal.
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Find the angle between the given vectors to the nearest tenth of a degree. u = &lt;-5, -4&gt;, v = &lt;-4, -3&gt; (1 point)
sesenic [268]

Angle between u = -5i-4j , v=-4i-3j is x =0° .

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

We have , two vectors u = <-5, -4>, v = <-4, -3>  or , u = -5i-4j , v=-4i-3j

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

u.v =|u|(|v|)cosx , where x is angle between u & v !

⇒ u.v =|u|(|v|)cosx

⇒ cosx =\frac{u.v}{|u|(|v|)}

Now , u.v = (-5i-4j)(-4i-3j)

⇒ u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)

⇒ u.v = 20+12            { i(j) = j(i) =0  }

⇒ u.v = 32

Now , Modulus of any vector  r = xi+yj is |r| = \sqrt{x^{2}+y^{2}} So ,

|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5

Putting all these values in equation cosx =\frac{u.v}{|u|(|v|)} we get:

⇒ cosx =\frac{32}{5(\sqrt{41})}

⇒ cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})

⇒ x =cos^{-1}(1)                 { cos0 = 1  }

⇒ x =0°

Therefore , Angle between u = -5i-4j , v=-4i-3j is x =0° .

8 0
3 years ago
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