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2 years ago

(will give brainliest) 2+5x-3x+6x

Mathematics

2 answers:

pickupchik [31]2 years ago

3 0

Answer:

x = -0.25

Step-by-step explanation:

2+5x-3x+6x

5x-3x+6x = -2

8x = -2

x = -2/8

x = -0.25

Igoryamba2 years ago

3 0

Answer: 2+8x

step by step solution

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Can someone please help me with this!!!!!<br><br> 2/5 ÷ 8 1/2 = ?

pav-90 [236]

The answer would be 4/85

1. 2/5 ÷ 8 * 2 + 1 over 2

2. 2/5 ÷ 16 + 1 over 2

3. 2/5 ÷ 17/2

4. 2/5 * 2/17

5. 2 * 2 over 5 * 17

6. 4 over 5 * 17

7. finally 4/85

7 0

2 years ago

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ASAP HELP!! PLS PLS ILL GIVE U BRAINLIST !! DUE IN 5 mins :(

Aliun [14]

I can’t see it could you type it and tell me what it says?

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2 years ago

What is the angle measure of 2x+1°

Ray Of Light [21]

Answer:

25 degrees

Step-by-step explanation:

Haha, awesome! Just did a problem like this!

All angles add up to 180, so set it up like this:

2x + 1 + 5x + 5 + 90 = 18

Combine like terms.

7x + 96 = 180

7x = 84

x = 12

Plug x back into 2x + 1

2(12) + 1

24 + 1

= 25 degrees

While we're at it...

5(12) + 5

60 + 5

= 65 degrees (for the other angle)

25 + 65 + 90 = 180

Math checks out & we're good to go!

Hope this helped. :) <3

5 0

2 years ago

This is a very good question that the brainly ppl cant delete, WHYYYYYYY?!

Arturiano [62]

Answer:

1000

Step-by-step explanation:

I search it it cam in a calculator

8 0

2 years ago

Read 2 more answers

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