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3 years ago

6

Why do teachers give out so much homework? I can understand why colledge and hugh school get it, but why the younger ones? They'

re still delevoping. Why? (This doesn't have to do with any subjects; it's not math)...

1 answer:

seraphim [82]3 years ago

8 0

Most teachers say that they do it to make sure the kids dont't forget what ws tought that day or to review

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A sum of money doubles itself in 16 2 years at certain rate of simple interest. How long will it take to double itself at the sa

Viefleur [7K]

Answer:

r=40% or .4

Step-by-step explanation:

32=16(1+r)^2

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3 years ago

0.04 rounded to the nearest hundreth<br><br> and <br><br> 0.2% rounded to the nearest hundreth

ELEN [110]

To round a number, you look to the next place to the right of what you want to round to... for example, if you want to round to the nearest hundredth, you look to the thousandths place to see whether the hundredths place rounds up or down.

0.04 is already at the hundredths place so the thousandths place is zero... the answer to the nearest hundredth is 0.04.

For 0.2%, you have to convert the percentage to a decimal by dividing by 100 (move the decimal 2 places to the left).

0.2% = 0.002

So now to round to the nearest hundredth, we look to the thousandths place. The thousandths place has a 2 in it so the hundredths place rounds down. 0.2% total he nearest hundredth is 0.

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3 years ago

What is the range and domain

mr_godi [17]

Answer:

domain is 0 to infinity and range is 5 to infinity

Step-by-step explanation:

x is domain and y is range

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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

I need help with question 12

Llana [10]

the answer is b.............................................

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3 years ago