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3 years ago

Fill in the blank: To obtain the graph of y=2x+0.2, you can start with the graph of y=2x and translate it ___________.

Mathematics

2 answers:

s344n2d4d5 [400]3 years ago

8 0

Answer:

y=2x+0.2, you can start with the graph of y=2x and translate it 0.2 units up .

Ad libitum [116K]3 years ago

4 0

Answer:

Step-by-step explanation:

+0.2 will increase every y-value by 0.2, hence shifting the graph 0.2 units upwards

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In a basketball game Fintan made 16 baskets. Each of the baskets is worth either 2 or 3

jekas [21]

Let x represent a 2-point basket, and y represent a 3-point basket.

Basket equation: x + y = 16
Point equation 2x + 3y = 39

Solve using system of equations:
if x + y = 16, then x = 16 - y

Plug into the second equation:
2(16-y) + 3y = 39
32 - 2y + 3y = 39
32 + y = 39
y = 7

So the number of 3-point baskets scored was 7. Hope this helps, let me know if there’s a step you don’t understand! :)

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3 years ago

The quotient obtained when an integer is divided by its additive inverse is ________________.

Ahat [919]

Answer:

-1

Step-by-step explanation:

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3 years ago

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Find the area of the shape shown below. Please answer.

dimulka [17.4K]

Answer:

Step-by-step explanation:

2x2=4/2=2 small triangle, 2x2=4 small square, 6x2/2=6 med. triangle, 6+2+2=10x4=40/2=20

2+4+6+20=32

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3 years ago

the names of nine boys and two girls are in a hat. what is the probability of first drawing a girls name and then a boys name fr

lara [203]

18/121 shoul be the answer

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3 years ago

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Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

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3 years ago