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2 years ago

Simplify 1+tan^2 x / tan^2 x a) sin^2 x b) csc^2 x c) sec^2 x d) cos^2 x

Mathematics

1 answer:

rodikova [14]2 years ago

5 0

Answer:

cosec²x is the answer

Step-by-step explanation:

$\frac{1+tan^2x}{tan^2x} \\\\\frac{sec^2x}{tan^2x} \\\\\frac{\frac{1}{cos^2x} }{\frac{sin^2x}{cos^2x} }\\\\\frac{1}{sin^2x}\\\\cosec^2x$

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Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

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3 years ago

The sum of two numbers is -14 one number is 20 less than the other find the numbers

Lisa [10]

Answer:

x=-17,y=3

Step-by-step explanation:

x+y=-14

x=x

y=x+20

x+(x+20)=-14

2x+20=-14

2x=-14-20

2x=-34

x=-34/2

x=-17

y=-17+20

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3 years ago

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Deffense [45]

Answer:

SOLVE IT UR SELF OR GET A CALCULATOR

Step-by-step explanation:

*urinates on ur property*

ANSWER : 225, 225, 225, 225, 225

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3 years ago

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3 years ago

What is the product of 4.5 × 10-5 and 2.4 × 10-2?

nikitadnepr [17]

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