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3 years ago

12

Why is it possible to have right equilateral triangles

2 answers:

dimulka [17.4K]3 years ago

7 0

Not possible because it has to be equal and if it had a right angle it wouldnt be equal

Paraphin [41]3 years ago

4 0

No it is not possible

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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

Differentiate y=x⁴(1-2x⁵)⁶(5-8x³)².

Liula [17]

Step-by-step explanation:

Let $y(x)=f(x)g(x)h(x)$ where

$f(x) = x^4$

$g(x)= (1 -2x^5)^6$

$h(x)= (5 - 8x^3)^2$

so that

$y(x) = x^4(1 -2x^5)^6(5 - 8x^3)^2$

Recall that the derivative of the product of functions is

$y'(x)=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$

so taking the derivatives of the individual functions, we get

$f'(x) = 4x^3$

$g'(x) = 6(1 - 2x^5)^5(-10x^4)$

$h'(x) = 2(5 - 8x^3)(-24x^2)$

So the derivative of y(x) is given by

$y'(x) = 4x^3(1 -2x^5)^6(5 - 8x^3)^2 + x^4 6(1 -2x^5)^5(-10x^4)(5 - 8x^3)^2 + x^4(1 -2x^5)^6 2(5 - 8x^3)(-24x^2)$

or

$y'(x) = 4x^3(1 -2x^5)^6(5 - 8x^3)^2$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 60x^8(1 -2x^5)^5(5 - 8x^3)^2$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 48x^6(1 -2x^5)^6 2(5 - 8x^3)$

3 0

3 years ago

What is ✓ 75 in simplified form?<br> O 5V5<br> 0 2573<br> O 3V 5<br> O 3V 25

faust18 [17]

It should be 5√3

Because 25 x 3 = 75

√25=5 so u bring that out and keep the 3 in because it’s a prime number

3 0

3 years ago

The most Stanley can afford to pay per year in mortgage payments is $12,000, and his credit score is currently 544. According to

iris [78.8K]

Answer:

131 points

Step-by-step explanation:

Apex. Sad nobody came in clutch for me

3 0

3 years ago

Read 2 more answers

Which expression is equivalent to ( 4 1/3)^3 ?

shutvik [7]

(13/3)^3 or (13/3)•(13/3)•(13/3)

5 0

2 years ago