Ask question

Ask question

All categories

2 years ago

12

Which of the following ordered pairs could be placed in the table and still have the relation qualify as a linear function? (4 p

oints)
Group of answer choices

(1, 7)

(−1, 13)

(2, 13)

(2, 7)

2 answers:

Sophie [7]2 years ago

8 0

Your answer would be (2,13)

Lerok [7]2 years ago

7 0

Answer:

(2, 7)

Step-by-step explanation:

You might be interested in

PLEASE HELP ME SEE IF MY ANSWERS ARE CORRECT!!!!!!!

Lena [83]

So first one
'how many solutions does 2x-y=-5 and 2x+y=5 have?'
add and get
2x-y-5
plus
2x+y=5

equals
2x+2x+y-y=5-5
4x=0
x=0 always
solve for y
4(0)+y=5
y=5
the solution is (0,5)
only <u>ONE </u>solution

one way is to subsitute
just remember that it is in (x,y) form so
the pont (1,2) means that 1 solution is x=1 and y=2 so subsitute and find that
the first one is the answer you are correct

just look at the graph
the solution is the intersection
it seems to be at a point that is 3 units to the right and -6 units up (6 units down)
so the solution is (3,-6)
yo are corect

subsitution
y=y
therefor
the answe ris (-4,-14) if you did the math correctly

#8 is correct

# 9 is correct

# 10 the answe ris bananas=0.40 pears=0.60

the last one you got it wrong, remember to check your answer to the graph for commonsense
then answer is (-2,5) and (1,2)

8 0

3 years ago

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

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

5 0

3 years ago

The ratio of fiction to nonfiction books in the library is 4:5. If there are 900 books in a collection, how many are fiction?

jekas [21]

There will be excactly 400 fiction books. The ratio of fiction to all is 4 to 9, or 4 to 900.

4 0

3 years ago

In the given equation, k is a constant. If the equation has no solution, what is the value of k?

MrRissso [65]

Answer:

0

Step-by-step explanation:

7 0

3 years ago

Forty is 40% less than what number

kirza4 [7]

40 = n-.40n

40 =n(1-.4)

40 = n(.6

divide by .6 on each side

40/.6 =n

n=66 2/3

5 0

3 years ago