What's does f(x) equal if I'm usin (-2,4) in the problem on that graph

Ann wants to choose from two telephone plans. Plan A involves a fixed charge of $10 per month and call charges at $0.10 per minu

Kay [80]

Ann wants to choose from two telephone plans. Plan A involves a fixed charge of $10 per month and call charges at $0.10 per minute. Plan B involves a fixed charge of $15 per month and call charges at $0.08 per minute.

Plan A $10 + .10/minute

Plan B $15 + .08/minute

If 250 minutes are used:

Plan A: $10+$25=$35
Plan B: $15+$20=$35

If 400 minutes are used:

Plan A: $10+$40=$50
Plan B: $15+$32=$47

B is the correct answer. How to test it:

Plan A: $10+(.10*249 minutes)
$10+$24.9=$34.9

Plan B: $15+(.08*249 minutes)
$15+$19.92=$34.92

Plan A < Plan B if less than 250 minutes are used.

4 0

3 years ago

How do you solve each system of equations?

stiv31 [10]

Answer:

$x=1,y=3,z=1$

Step-by-step explanation:

$4x-4y+4z=-4$

$4x=-4+4y-4z$

$x=-1+y-z$

Substitute $x=-1+y-z$ into second equation:

$4\left(-1+y-z\right)+y-2z=5$

$-4+4y-4z+y-2z=5$

$5y-6z-4=5$

$y=\frac{6z+9}{5}$

Substitute $x=-1+y-z$ and $y=\frac{6z+9}{5}$ into the third equation:

$\frac{-41z-54}{5}+3=-16$

$-41z-54=-95$

$z=1$

Substitute $z=1$ into $y=\frac{6z+9}{5}$ :

$y=3$

Plug in y and z values into $x=-1+y-z$ :

$x=1$

4 0

4 years ago

Use your knowledge on the triangle proportionality theorem to find the missing side length indicated. Be sure to explain the pro

ikadub [295]

Answer:

? = 4.29

Step-by-step explanation:

Remark

Let x be the question mark. You get the proportionality by using the dimensions of the little triangle to the dimensions of the large triangle.

Equation

14/(14+ 4) = 15/(x + 15) Combine like terms on the left

14/18 = 15/(x + 15) Cross multiply

14*(x + 15) = 18 * 15 Simplify the right

14(x + 15) = 270 Divide by 14

x + 15 = 270 / 14

x + 15 = 19.29 Subtract 15 from both sides

x = 19.29 - 15

x = 4.29

6 0

3 years ago

Use the given transformation to evaluate the given integral, where r is the triangular region with vertices (0, 0), (8, 1), and

Jlenok [28]

We first obtain the equation of the lines bounding R.

For the line with points (0, 0) and (8, 1), the equation is given by:

$\frac{y}{x} = \frac{1}{8} \\ \\ \Rightarrow x=8y \\ \\ \Rightarrow8u+v=8(u+8v)=8u+64v \\ \\ \Rightarrow v=0$

For the line with points (0, 0) and (1, 8), the equation is given by:

$\frac{y}{x} = \frac{8}{1} \\ \\ \Rightarrow y=8x \\ \\ \Rightarrow u+8v=8(8u+v)=64u+8v \\ \\ \Rightarrow u=0$

For the line with points (8, 1) and (1, 8), the equation is given by:

$\frac{y-1}{x-8} = \frac{8-1}{1-8} = \frac{7}{-7} =-1 \\ \\ \Rightarrow y-1=-x+8 \\ \\ \Rightarrow y=-x+9 \\ \\ \Rightarrow u+8v=-8u-v+9 \\ \\ \Rightarrow u=1-v$

The Jacobian determinant is given by

$\left|\begin{array}{cc} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{array}\right| = \left|\begin{array}{cc} 8 &1\\1&8\end{array}\right| \\ \\ =64-1=63$

The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v

Therefore, the integration is given by:

$63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576$

6 0

3 years ago

Please show work!!!!!!

Ksivusya [100]

Answer:

3×30000×10/100

3×300×10

30×300

9000

7 0

3 years ago

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