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3 years ago

How do i answer this? PLEASE HELP

Mathematics

1 answer:

scZoUnD [109]3 years ago

4 0

The teacher circled the answer for u. so i dont see what u need help with.

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PLSS ANSWER

labwork [276]

$\frac{3}{6} = three \: sixths$

$\frac{3}{6} + \frac{3}{6} = \frac{3 + 3}{6} = \frac{6}{6} = 1$

<h2>Letter C</h2>

8 0

3 years ago

Identify the terms and like terms 10x + 5 + 3x + 1

Svetach [21]

Answer:

10x +3x are like terms

5+1 is terms

Step-by-step explanation:

8 0

3 years ago

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How do I solve this?

ozzi

The answer is A) √3
cot(-5π/6)=1/tan(-5π/6)
=1/tan(-(5π/6)
Use the property: tan(-x)=-tan(x)
=1/tan(5π/6)
-(1/tan(5π/6))
tan(5π/6)=-√3/3
-(1/-√3/3)
-(1/-√3/3)
=√3

4 0

3 years ago

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find the values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (1, -8)

frutty [35]

Answer:

cosФ = $\frac{1}{\sqrt{65}}$ , sinФ = $-\frac{8}{\sqrt{65}}$ , tanФ = -8, secФ = $\sqrt{65}$ , cscФ = $-\frac{\sqrt{65}}{8}$ , cotФ = $-\frac{1}{8}$

Step-by-step explanation:

If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:

cosФ = $\frac{x}{r}$
sinФ = $\frac{y}{r}$
tanФ = $\frac{y}{x}$
secФ = $\frac{r}{x}$
cscФ = $\frac{r}{y}$
cotФ = $\frac{x}{y}$

Where r = $\sqrt{x^{2}+y^{2} }$ (the length of the terminal side from the origin to point (x, y)

You should find the quadrant of (x, y) to adjust the sign of each function

∵ Point (1, -8) lies on the terminal side of angle Ф in standard position

∵ x is positive and y is negative

→ That means the point lies on the 4th quadrant

∴ Angle Ф is on the 4th quadrant

∵ In the 4th quadrant cosФ and secФ only have positive values

∴ sinФ, secФ, tanФ, and cotФ have negative values

→ let us find r

∵ r = $\sqrt{x^{2}+y^{2} }$

∵ x = 1 and y = -8

∴ r = $\sqrt{x} \sqrt{(1)^{2}+(-8)^{2}}=\sqrt{1+64}=\sqrt{65}$

→ Use the rules above to find the six trigonometric functions of Ф

∵ cosФ = $\frac{x}{r}$

∴ cosФ = $\frac{1}{\sqrt{65}}$

∵ sinФ = $\frac{y}{r}$

∴ sinФ = $-\frac{8}{\sqrt{65}}$

∵ tanФ = $\frac{y}{x}$

∴ tanФ = $-\frac{8}{1}$ = -8

∵ secФ = $\frac{r}{x}$

∴ secФ = $\frac{\sqrt{65}}{1}$ = $\sqrt{65}$

∵ cscФ = $\frac{r}{y}$

∴ cscФ = $-\frac{\sqrt{65}}{8}$

∵ cotФ = $\frac{x}{y}$

∴ cotФ = $-\frac{1}{8}$

8 0

3 years ago

A pound of cherries is normally $5.00. Today there is a 20% discount. What is the sale price

Kruka [31]

The Answer Is $4.00
Hope This Helps!

6 0

4 years ago

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