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2 years ago

7

Which statement is true about the slope of the graphed line?

1 answer:

Sonbull [250]2 years ago

4 0

B. the slope is positive because its in a straight line

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Rico spent $87.00 to go to the movies on Friday night. He spent $25.75 on popcorn and drinks and paid $12.25 for each movie tick

Svetlanka [38]

You haven't given the equations I'm supposed to chose from, but the answer would be

25.72 + 12.25x = 87.00 is the equation.

He spent a total of $87

The popcorn was 25.75 and each ticket was 12.25, which means we must add the amounts they costs and multiply how my tickets he bought, which is x.

5 is the solution

7 0

2 years ago

Read 2 more answers

Complete the square 3x^2+12x=15

barxatty [35]

Step-by-step explanation:

$3 {x}^{2} + 12x = 15 \\ 3 ({x}^{2} + 4x) = 15 \\ {x}^{2} + 4x = 5 \\ {x}^{2} + 4x = 5 \\ {x}^{2} + 4x + 4 = 5 + 4 \\ {x}^{2} + 4x + 4 = 9 \\ {x}^{2} + 4x + {2}^{2} = {3}^{2} \\ \red{ \bold{ (x + 2)^{2} = {3}^{2}}} \\$

3 0

3 years ago

Please tell me what this is

Harlamova29_29 [7]

Answer:7.2+z=10

Step-by-step explanation:if I am right you just have to do the equacion

4 0

2 years ago

What is -3/5 x -10/21?

sammy [17]

The answer would be 2/7!

7 0

3 years ago

Determine whether the given differential equation is exact. If it is exact, solve it. (If it is not exact, enter NOT.) (tan(x) −

never [62]

Answer:

$f(x,y)=ln secx+cosx siny+C$

Step-by-step explanation:

We are given that DE

$(tanx-sinx siny)dx+cosxcosydy=0$

We have to determine given DE is exact or not.

Compare it with Mdx+Ndy=0

$M=tanx-sinx siny$

$N=cosxcosy$

$\frac{\partial M}{\partial y}=$ $M_y=-sinxcosy$

$\frac{\partial N}{\partial x}=N_x=-sinxcosy$

Therefore, $M_y=N_x$

If DE is exact then $M_y=N_x$

Hence,it is exact.

$M=\frac{\partial f}{\partial x}=tanx-sinxsiny$

Integrate w.r.t x on both sides

$f(x,y)=\int(tanx-sinxsiny)dx$

$f(x,y)=lnsecx+cosxsiny+\phi(y)$ ...(1)

By using the formula

$\int tanxdx=lnsecx+C,\int sinxdx=-cosx+C$

Differentiate partially equation (1) w.r.t y

$\frac{\partial f}{\partial y}=cosxcosy+\phi'(y)$

By using the formula:

$\frac{d(sinx)}{dx}=cosx$

$N=\frac{\partial f}{\partial y}=cosxcosy=cosxcosy+\phi'(y)$

$\phi'(y)=cosxcosy-cosxcosy=0$

$\phi'(y)=0$

Integrate w.r.t y

$\phi(y)=C$

Substitute the value

$f(x,y)=ln secx+cosx siny+C$

7 0

3 years ago