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Minchanka [31]
3 years ago
8

Look at shape III in graph 2. Is there a dilation that maps shape Il onto shape III? If so, what is the scale factor and is it a

n enlargement or a
reduction?

Mathematics
1 answer:
Semmy [17]3 years ago
3 0

Answer:

The shape is a reduction, and the scale factor is actually 2/3. I was wrong but I have the correct answer thanks to a math teacher.

Step-by-step explanation:

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Is y =(-2)* an exponential function? Justify your answer.​
zimovet [89]

Answer:

But it's not an exponential function. In an exponential function, the independent variable, or x-value, is the exponent, while the base is a constant. For example, y = 2x would be an exponential function.

Step-by-step explanation:

the normal exponential function is ab^x, where a represent the growth( sometimes you don't need this part), b is average rate of growth and x is the number of times we exponate.  is a exponetital function because it includes the b and x part. The a part is someetimes used but not always.

5 0
3 years ago
I am confused can somebody help me<br> F ( - 5 ) =
vagabundo [1.1K]

Answer:

-5f

Step-by-step explanation:

3 0
2 years ago
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Find the median and mean of the data set below:<br> 47,15, 6, 49, 45, 30
Georgia [21]

Answer:

The median of the data set:

6, 15, 30, 45, 47, 49

The middle numbers are 30 and 45 so

You have to do 30 + 45

= 75

Then do 75 ÷ 2

= 37.5

The mean of the data set:

47, 15, 6, 49, 45, 30 (add them all)

= 192

Then divide 192 by 6 (because there are 6 numbers)

192 ÷ 6

= 32

So the median is 37.5 and the mean is 32.

Step-by-step explanation:

Hope this helps!

From your neighborhood softie :)

5 0
2 years ago
Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

\implies F(x,y)=x^2y^3+\ln(x+y^2)=C

With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
3 years ago
Quick TIMED does anyone know this algebra 2 question? Will give brainiest
USPshnik [31]

I think it is the third one

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3 years ago
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