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2 years ago

Eventhough i solved it i don't understand how to fill the colors

Mathematics

2 answers:

Lisa [10]2 years ago

7 0

"color the piece according to your color chart." Do you have that? If you don't, I wouldn't worry, you still did all the work and got all the answers, so your teacher should be fine with it.

Rainbow [258]2 years ago

6 0

If you weren't assigned a specific color for each box, choose a color for the four rectangles. Each answer in the section should be colored with that same color. It looks like you have that golden yellow for everything in the lower left section.

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A dot plot titled hours of practice this week going from 0 to 4. 0 has 1 dot, 1 has 2 dots, 2 has 0 dots, 3 has 4 dots, and 4 ha

WITCHER [35]

Answer:

1, 2, 3, 5

Step-by-step explanation:

I took the quiz

4 0

3 years ago

What is the rate of change of y with respect to x for this function?

lilavasa [31]

Answer:

Given the function: y=f(x) = 3x+2

when x=-2 at the beginning of the interval [-2, 5],

then;

y = 3x+2 begins at

y= 3(-2)+2 = -6+2= -4.

and

when x=5 at the end of the interval [-2, 5],

y = 3x+2 ends up at

y= 3(5)+2 = 15+2= 17.

So,

y has changed -4 to 17, which is a change of 17-(-4)= 17+4 = 21

and x has changed from -2 to 5, which is a change of 5-(-2)=5+2=7

So, the average rate of change of y with respect to x over the interval

[-2, 5] is given by ;

Therefore, the average rate of change y with respect to x over the interval is, 3

Step-by-step explanation:

4 0

2 years ago

Are any concentric circles shown?<br><br> yes<br> no

dimulka [17.4K]

No because there is no picture and circles

5 0

3 years ago

Read 2 more answers

Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (−1, 2) to (1, 2) (a) find a function f

Korvikt [17]

If there is some scalar function $f(x,y)$ such that

$\nabla f(x,y)=\mathbf f(x,y)=x^2\,\mathbf i+y^2\,\mathbf j$

then we want to find $f$ such that

$\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)$
$\dfrac{\partial f}{\partial y}=y^2=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\dfrac{y^3}3+C$
$\implies f(x,y)=\dfrac{x^3}3+\dfrac{y^3}3+C$

So the vector field $\mathbf f(x,y)$ is conservative, which means the fundamental theorem applies; the line integral of $\mathbf f$ along any path $\mathcal C$ parameterized by some vector-valued function $\mathbf r(t)$ over $a\le t\le b$ is given by

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=a}^{t=b}\mathbf f(\mathbf r(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt=f(\mathbf r(b))-f(\mathbf r(a))$

In this case,

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(1,2)-f(-1,2)=\dfrac23$

5 0

2 years ago

Solve the equation for x by graphing.

juin [17]

D is your answer!!!!

8 0

3 years ago