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seropon [69]
3 years ago
8

I NEED HELP PLEASE, THANKS! :)

Mathematics
2 answers:
Afina-wow [57]3 years ago
6 0

Answer:

Third option is the right choice.

Step-by-step explanation:

Manipulate the right hand side.

\sec \left(x\right)\\\\=\frac{1}{\cos \left(x\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \:1=\cos ^2\left(x\right)+\sin ^2\left(x\right)\\\\=\frac{\cos ^2\left(x\right)+\sin ^2\left(x\right)}{\cos \left(x\right)}\\\\\Rightarrow \mathrm{True}

Best Regards!

Andreyy893 years ago
6 0

Answer:

sin ^2 (x)+ cos^2(x)  = 1

Step-by-step explanation:

sin ^2 (x)+ cos^2(x)

------------------------

cos (x)

We know that  sin ^2 (x)+ cos^2(x)  = 1

1

------------------------

cos (x)

The definition of sec is 1 /cos

= sec(x)

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Answer:

4.975 meters

Step-by-step explanation:

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4.975 meters

-Chetan K

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3 years ago
Find the square. (1/4A + 1/4B)^2<br><br> a) 1/16A^2 + 1/8AB + 1/16B^2<br> b) 1/4A^2 + 1/8AB + 1/4B^2
alisha [4.7K]

(a + b)^2 = a^2 + 2ab + b^2

(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =

= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B^2

Answer: A.

Another way to solve by factoring 1/4.

(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =

= [\dfrac{1}{4}(A + B)]^2

= (\dfrac{1}{4})^2(A + B)^2

= \dfrac{1}{16}(A^2 + 2AB + B^2)

= \dfrac{1}{16}A^2 + \dfrac{1}{16}2AB + \dfrac{1}{16}B)^2

= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B)^2

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3 years ago
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never [62]
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If f(x) = 2x^4, then f'(3) =
valkas [14]

To find f'(3) (f prime of 3), you must find f' first.  f' is the derivative of the function f(x).

Finding the derivative of f(x) = 2x⁴ requires the use of the power rule.

The power rule for derivatives is \frac{d}{dx} [x^{n} ] =nx^{n-1}.  In other words, you bring the exponent forward and multiply it by the coefficient of the term, and then you subtract 1 from the original exponent.

f'(x) = \frac{d}{dx}(2x⁴)

f'(x) = 2(4)x³

f'(x) = 8x³

Now, to find f'(3), plug 3 into your derivative.

f'(3) = 8(3)³

f'(3) = 216

<h3>Answer:</h3>

f'(3) = 216

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3 years ago
Secants P N and L N intersect at point N outside of the circle. Secant P N intersects the circle at point Q and secant L N inter
JulijaS [17]

Answer:

A circle is shown. Secants P N and L N intersect at point N outside of the circle. Secant P N intersects the circle at point Q and secant L N intersects the circle at point M. The length of P N is 32, the length of Q N is x, the length of L M is 22, and the length of M N is 14.

In the diagram, the length of the external portion of the secant segment PN is <u>X</u>

The length of the entire secant segment LN is <u>36</u>.

The value of x is <u>15.74</u>

Step-by-step explanation:

Snap

Jona_Fl16

7 0
3 years ago
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