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2 years ago

In the definition of fractional numbers, b ≠ 0. Why?

Mathematics

1 answer:

Nataliya [291]2 years ago

6 0

I have no idea. Just commenting to get the answer when someone else knows it

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Verify that this trigonometric equation is an identity?

Vlada [557]

$\cot x\sec^4 x=\cot x+2\tan x+\tan^3x\\\\L=\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\cos^4x}=\dfrac{1}{\sin x}\cdot\dfrac{1}{\cos^3x}=\dfrac{1}{\sin x\cos^3x}\\\\R=\dfrac{\cos x}{\sin x}+2\cdot\dfrac{\sin x}{\cos x}+\dfrac{\sin^3x}{\cos^3x}\\\\=\dfrac{\cos x\cos^3x}{\sin x\cos^3x}+\dfrac{2\sin x\cos^2x}{\cos x\sin x\cos^2x}+\dfrac{\sin^3x\sin x}{\cos^3x\sin x}\\\\=\dfrac{\cos^4x+2\sin^2x\cos^2x+\sin^4x}{\sin x\cos^3x}\\\\=\dfrac{(\cos^2x)^2+2\sin^2x\cos^2x+(\sin^2x)^2}{\sin x\cos^3x}$

$=\dfrac{(\cos^2x+\sin^2x)^2}{\sin x\cos^3x}=\dfrac{1}{\sin x\cos^3x}=L\\\\Used:\\\tan(a)=\dfrac{\sin(a)}{\cos(a)}\\\cot(a)=\dfrac{\cos(a)}{\sin(a)}\\\sec(a)=\dfrac{1}{\cos(a)}\\\sin^2a+\cos^2a=1\\(a+b)^2=a^2+2ab+b^2$

8 0

3 years ago

Describe the graph that is produced by the equation (x-7)^2+(y+5)^2>25

Alborosie

Step-by-step explanation:

The equation will form a circle because it in the form of

$(x - h) {}^{2} + (y - k) {}^{2} = {r}^{2}$

First, let set the equation equal.

$(x - 7) {}^{2} + (y + 5) {}^{2} = 25$

Here the center will be (7,-5), and the radius of 5. The boundary line will be dashed

Since this is in the inequalities, we must find the solution set.

Plug in 0,0 for x and y and see if it's true.

$(0 - 7) {}^{2} + (0 + 5) {}^{2} > 25$

$49 + 25 > 25$

$74 > 25$

This is a true so we shade the region that includes 0,0

Since 0,0 has a greater distance from the center of the circle, 0,0 is outside of the circle, so our solution set will

be outside of circle.

Here a picture of graph,

8 0

2 years ago

Eugene has five payments left to make on his computer. If each payment is $157.90, how much does he still owe?

puteri [66]

$157.90 x 5 = $789.50

He still owes $789.50

3 0

2 years ago

Angelina received two discounts on a $50 pair of shoes. The discounts were taken off one after the other. Angelina used a 15% co

Lemur [1.5K]

Answer:

$40

Step-by-step explanation:

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=y%3D1-sqrt%281-2%2Fx%29" id="TexFormula1" title="y=1-sqrt(1-2/x)" alt="y=1-sqrt(1-2/x)" align=

Svetllana [295]

$~~~~~~~~y = 1- \sqrt{ 1- \dfrac 2x}\\\\\\\implies \sqrt{ 1- \dfrac 2x} = 1 - y\\\\\\\implies 1 - \dfrac 2x = (1 -y)^2~~~~~~~~~~~~~;[\text{Square on both sides}]\\\\\\\implies \dfrac 2x = 1-(1-y)^2\\\\\\\implies \dfrac{1}{x} = \dfrac{1-(1-2y+y^2)}{2}\\\\\\\implies x = \dfrac{2}{1-1+2y-y^2}\\\\\\\implies x = \dfrac{2}{2y-y^2}\\\\\\\implies x = -\dfrac{2}{y(y-2)}~~~~~~~~~~~~~~~~~;[y\neq 0,~ y \neq 2]\\$

6 0

2 years ago