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3 years ago

Simplify and write the trigonometric expression in terms of sine and cosine:

Mathematics

1 answer:

kicyunya [14]3 years ago

6 0

Answer:

f(x) = + cosx or - cosx

Step-by-step explanation:

(2+tan^2x/sec^2x) -1 =(f(x))^2

=(( 1 + sec^2x)/sec^2x) -1

= 1 / sec^2x + 1 - 1

= 1 / sec^2x

= cos^2x

(f(x))^2 = cos^2x

f(x) = + cosx or - cosx

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Y=-x^2+2x+10 y=x+2 Substitution Need ASAP

aniked [119]

Answer:

$x_1=\frac{1+\sqrt{33} }{2},y_1=\frac{5+\sqrt{33} }{2}\\ \\ x_2=\frac{1-\sqrt{33} }{2},y_2=\frac{5-\sqrt{33} }{2}$

Explanation:

You must find the solution of the system using the substitution method.

System:

$y=-x^2+2x+10\\\\ y=x+2$

Substitute x+2 for y:

$x+2=y=-x^2+2x+10$

Pass all the terms to one side:

$x^2-2x+x+2-10\\ \\ x^2-x-8=0$

Use the quadratic formula to solve:

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}\\ \\ x=\frac{-(-1)+/-\sqrt{(-1)^2-4(1)(-8)} }{2(1)}\\ \\ x=\frac{1+/-\sqrt{1+32} }{2}$

$x_1=\frac{1+\sqrt{33} }{2}\\ \\x_2=\frac{1-\sqrt{33} }{2}$

Use y = x + 2 to find the value of y:

$y_1=2+x_1=2+\frac{1+\sqrt{33} }{2}\\ \\y_1=\frac{5+\sqrt{33} }{2} \\\\y_2=2+x_2=2+\frac{1-\sqrt{33} }{2}\\ \\ y_2=\frac{5-\sqrt{33} }{2}$

6 0

3 years ago

Are the ratios 13:14 and 1:2 equivalent?

eduard

Answer:

NO!

Step-by-step explanation:

They are not equivalent. We can check this. If you rewrite them as fractions 13/14 and 1/2, you can cross multiply and see if they are equal. Which they are not. Also, you could just guess.... think of 13:14 as 13 out of 14. And 1:2 as 1 out of 2. You can tell that 13/14 is much greater than 1/2.

6 0

3 years ago

Pentagon ABCDE is shown on the coordinate plane below

77julia77 [94]

Answer: c) B' = (6, -3)

Step-by-step explanation:

The simplest way is to rotate the graph.

Rotating the graph 1 time to the left (counterclockwise) = 90°

Rotating the graph 2 times to the left (counterclockwise) = 180°

Rotating the graph 3 times to the left (counterclockwise) = 270°

The longer way is to reflect over the x-axis and over the y-axis, which is the same as a rotation of 180°

Z = (x, y) → Z' = (-x, -y)