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Tamiku [17]
3 years ago
14

Look at the graph .which is the relationship between x and y PLSSS HURRY (100 POINTS

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
6 0

Answer:

it is B

Step-by-step explanation:

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Is this the same question from the other one.
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Find the value of x 117,145,133,140,119,125
andreev551 [17]

Answer:

here u go hope this is right

Step-by-step explanation:

x + 779 = 900

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What is the question of the line that passes through the point (-2,-2) and has a slope of 4. Plz someone answer
Studentka2010 [4]

Answer:y=4x+6

Step-by-step explanation:

We have the information that the slope is 4 and the line goes through the point (-2,-2). With this information, we can make a linear equation in a point slope form (y - y_{1}= m * (x - x_{1}), so the equation would be y+2=4(x+2), or simplified, y+2=4x+8. in order to solve for y (to make it a slope-intercept equation), we must subtract 2 from both sides. This gives us the equation y=4x+6. Hope this helps!

4 0
3 years ago
Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun
WITCHER [35]

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

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3 years ago
Need help with this data table need it to end up to 1024 thanks
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