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3 years ago

14

Im starting to calm down so thank you too everyone answering❤️

2 answers:

jeka57 [31]3 years ago

6 0

Answer:

E: 3 11/15

Step-by-step explanation:

Neko [114]3 years ago

4 0

Answer:

E

Step-by-step explanatio

its right

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Can someone help me?

mestny [16]

Answer:

Step-by-step explanation:

To do this, we must use the unit circle. At the angle $\frac{\pi }{6}$ the coordinate pair is ( $\frac{\sqrt{3} }{2}$ , $\frac{1}{2}$ ) and the hypotenuse is 1

Now that we know these we can start using the trig functions

sin = y

cos = x

tan = $\frac{y}{x}$

And the other functions would be the reciprocal of these

csc = $\frac{1}{y}$

sec = $\frac{1}{x}$

cot = $\frac{x}{y}$

This would mean that

sinx = $\frac{1}{2}$

cosx = $\frac{\sqrt{3} }{2}$

tanx = $\frac{1}{\sqrt{3} }$ or $\frac{\sqrt{3} }{3}$ if you're supposed to rationalize your denominators

cscx = 2

secx = $\frac{2}{\sqrt{3} }$ or $\frac{2\sqrt{3} }{3}$

cot = $\sqrt{3}$

6 0

2 years ago

Express 83% as a decimal. *<br> 8.3<br> 0.83<br> 83<br> 0.083

svetlana [45]

Answer:

83% would be 0.83 as a decimal

8 0

3 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago

Elijah and Jonathan play on the same soccer team. They have played 3 of their 15 games. They each create a model to represent x,

AlladinOne [14]

what are the optionsss??

7 0

2 years ago

Simplify the expression (4x - 3)(x + 5).

Drupady [299]

(4x-3)(x+5)
= 4x^2+20x-3x-15
= 4x^2+ 17x -15.

The final answer is 4x^2+ 17x -15~

3 0

3 years ago