All categories

2 years ago

Your question

Mathematics

1 answer:

Alex777 [14]2 years ago

8 0

2x+4.7=77.8

double the unknown number then add 4.7

You might be interested in

Find the cost for junior tickets

Marat540 [252]

what is the cost for? ik it's for juniors but what is the topic

4 0

3 years ago

Which statements are true about the figure? Select two

Nat2105 [25]

Answer:

The correct answers are A, C, D!

Step-by-step explanation:

I hope you do great!!

4 0

3 years ago

To avoid modeling the physical characteristics of wheel encoders such the square wave outputs, it is desirable to generate encod

grandymaker [24]

Answer:

The probability distribution is Normal continuous

Step-by-step explanation:

The basic idea is that velocity values have different noise level and an important thing regarding continuous probability distributions is that the probability of the random variable is equal to a specific outcome is 0

In other words, is practically impossible that one value of velocity could be the same as others.

5 0

2 years ago

Use x=5, equals, 5 to identify the value of each expression.

Lesechka [4]

Answer:

The answer to the first question is 625

The answer to the second question is 1

The answer to the third question is 1

4 0

3 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