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4 years ago

Which graph represents the solution set to this system of equations? Y=-1/2x+3 and y= 1/2x-1

Mathematics

2 answers:

Juliette [100K]4 years ago

7 0

Hey there, I hope I can be of assistance today!
While there are no graphs, I can provide you with the graph to answer your question assuming you have choices of graphs!

Hope this helps!

Bingel [31]4 years ago

4 0

Answer:

The graph is attached.

Step-by-step explanation:

We have been given a system of two equations. We are asked to find the graph that represents the solution set to this system of equations.

$y=-\frac{1}{2}x+3...(1)$

$y=\frac{1}{2}x-1...(2)$

First of all, we will find the solution of both equations by equating them as:

$\frac{1}{2}x-1=-\frac{1}{2}x+3$

$\frac{1}{2}x+\frac{1}{2}x-1=-\frac{1}{2}x+\frac{1}{2}x+3$

$\frac{1+1}{2}x-1=3$

$\frac{2}{2}x-1=3$

$x-1=3$

$x-1+1=3+1$

$x=4$

Upon substituting $x=4$ in equation (1), we will get:

$y=-\frac{1}{2}(4)+3$

$y=-2+3$

$y=1$

Since the solution of our given system is $(4,1)$ , therefore, the both graphs will intersect at $(4,1)$ .

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4 years ago

Read 2 more answers

Use calculus to find the absolute maximum and minimum values of the function. (round all answers to three decimal places.) f(x)

Allisa [31]

Part A:

Given the function $f(x)=x+2\cos(x)$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6}$

Using the second derivative test,

$f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732$

Since the second derivative gives a negative number, the given function has a maximum point at $x=\frac{\pi}{6}$ .

And the maximum point is given by:

$f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}$

i.e. $\left(\frac{\pi}{6},\ 2.256\right)$

Part B:

Given the function $f(x)=e^{-x}-e^{-2x}$ , the absolute maximum or minimum occurs when $f'(x)=0$ .

$f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931$

Using the second derivative test,

$f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5$

Since the second derivative gives a negative number, the given function has a maximum point at $x=0.6931$ .

And the maximum point is given by:

$f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}$

i.e. (0.693, 0.25)

3 0

3 years ago

Solve this quadratic equation using the quadratic formula. 5 - 10x - 3x2 = 0

horsena [70]

Answer:

x = (-5 ± 2√10) / 3

Step-by-step explanation:

5 − 10x − 3x² = 0

Write in standard form:

-3x² − 10x + 5 = 0

Solve with quadratic formula:

x = [ -b ± √(b² − 4ac) ] / 2a