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3 years ago

Solve |2x - 5| = 4 Thank you

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

8 0

Answer: x= 2 /9

Step-by-step explanation:

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Solve the system of equations. −2x+5y =−35 7x+2y =25

Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

$\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}$

This system of equations can be solved in three different ways:

Graphing the equations (method used)
Substituting values into the equations
Eliminating variables from the equations

Graphing the Equations

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is $\text{y = mx + b}$ .

Equation 1 is $-2x+5y = -35$ . We need to isolate y.

$\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7$

Equation 1 is now $y=\frac{2}{5}x-7$ .

Equation 2 also needs y to be isolated.

$\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}$

Equation 2 is now $y=-\frac{7}{2}x+\frac{25}{2}$ .

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}$

$\bullet \ \text{For x = 0,}$

$\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5$

Now, we can place these values in our table.

As we can see in our table, the rate of decrease is $-\frac{2}{5}$ . In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract $-\frac{2}{5}$ from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be $y=-\frac{7}{2}x+\frac{25}{2}$ . Therefore, we just use the same process as before to solve for the values.

$\bullet \ \text{For x = 0,}$

$\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5$

And now, we place these values into the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1 Equation 2

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$ $\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

Therefore, using this data, we have one solution at (5, -5).

4 0

3 years ago

Help me with thissss too

kramer

Answer:

dont know

Step-by-step explanation:

6 0

2 years ago

What is meant mathematically when we say we have an "even" or "odd" function?

Klio2033 [76]

I a function is even the leading term will have an even exponent. If a function is odd than the leading terms exponent is odd.

an even function has a parabola looking graph, while an odd function has min and max graph

6 0

3 years ago

A.Find the measures of the remaining angles of parallelogram KLMN. Justify each of the measures with a theorem.

Inga [223]

<K = <M = 50 ...Opposite angels are congruent

<K + <L = 180...Consecutive angles are supplementary

50 + <L = 180 so <L = 180 - 50 = 130

<L = <N = 130...Opposite angels are congruent

50 + <N = 180 so <N = 180 - 50 = 130

MN = LK = 7...Opposite sides are congruent

KN = LM = 10...Opposite sides are congruent

4 0

3 years ago

Simplify 3/5+(-1/4) 7/10

Vadim26 [7]

Answer:

17/40

Step-by-step explanation:

8 0

3 years ago

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