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Black_prince [1.1K]
3 years ago
11

Need help asap!! (summer packet)

Mathematics
1 answer:
DiKsa [7]3 years ago
6 0

Part A:

Function A:

Slope = (7-3)/(5-3) = 4/2 = 2

Equation:

y - 3 = 2(x - 3)

y - 3 = 2x - 6

y = 2x - 3

Funcion B:

(0, 3 ) and (-5, 0)

Slope = (3 - 0)/(0 + 5) = 3/5

y-intercept (0,3) so b = 3

Equation:

y = 3/5 x + 3

Function C:

y = 3x + 1

Part B:

Rate of change is the change in y over the change in x (rise/run). It's also the slope

Function A:     rate of change = 2

Function B:     rate of change = 3/5 (smallest)

Function C:     rate of change = 3   (largest)

Order linear functions based on rate of change from least to greatest.

Function B:             y = 3/5 x + 3

Function A:             y = 2x - 3

Function C:             y = 3x + 1

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PIT_PIT [208]
The <u>correct answers</u> are:

The <span><u>first system</u> </span>matches with (-4, 5);
The <u>second system</u> matches with (2, 2); 
The <u>third sys</u><span><u>tem</u> </span>matches with (-1, -3);
The <u>fourth system</u> matches with (3, 4);
The <u>fifth system</u> matches with (2, -1); and 
<span>The <u>sixth system</u> matches with (-5, 6).</span>

<span>Explanation:</span>

The <u>sixth system</u> is the easiest one to find.  We are given the value of x and the value of y in the equations: x=-5 and y=6.

For all other systems, we must solve the system.  For the <u>first system</u>:
\left \{ {{2x-y=-13} \atop {y=x+9}} \right.

Since we have the y-variable isolated in the second equation, we will use substitution.  We substitute this value in place of y in the first equation:
2x-y=-13
2x-(x+9)=-13

Distributing the subtraction sign,
2x-x-9=-13

Combining like terms:
x-9=-13

Add 9 to each side:
x-9+9=-13+9
x=-4

Substitute this into the second equation:
y=x+9
y=-4+9
y=5

<u>The coordinates (-4, 5) represent the solution point.</u>

For the <u>second system</u>:
\left \{ {{3x+2y=10} \atop {6x-y=10}} \right.

We can make the coefficients of x the same and use elimination.  To do this, we will multiply the first equation by 2:
<span>\left \{ {{2(3x+2y=10)} \atop {6x-y=10}} \right. \\ \\ \left \{ {{6x+4y=20}  \atop {6x-y=10}} \right.

Since the coefficients of x are now the same, we can cancel it.  They are both positive, so we subtract:
\left \{ {{6x+4y=20} \atop {-(6x-y=10)}} \right. \\ \\5y=10

Divide both sides by 5:
5y/5=10/5
y=2

Substitute this into the second equation
6x-y=10
6x-2=10

Add 2 to each side:
6x-2+2=10+2
6x=12

Divide both sides by 6:
6x/6 = 12/6
x=2

<span><u>The coordinates (2, 2) represent the solution to this system.</u></span>

<span>For the <u>third system</u>:</span>
\left \{ {{4x-3y=5} \atop {3x+2y=-9}} \right.

We can make the coefficients of x the same by multiplying the first equation by 3 and the second by 4:
\left \{ {{3(4x-3y=5)} \atop {4(3x+2y=-9)}} \right. \\ \\ \left \{ {{12x-9y=15} \atop {12x+8y=-36}} \right.

Since the coefficients of x are the same, we can cancel them.  Since they are both positive, we will subtract:
\left \{ {{12x-9y=15} \atop {-(12x+8y=-36)}} \right. \\ \\-17y=51

Divide both sides by -17:
-17y/-17 = 51/-17
y=-3

Substitute this into the first equation:
4x-3y=5
4x-3(-3)=5
4x--9=5
4x+9=5

Subtract 9 from each side:
4x+9-9=5-9
4x=-4

Divide each side by 4:
4x/4 = -4/4
x=-1

<u>The coordinates (-1, -3) represent the solution of the third system..</u>

For the <u>fourth system</u>:
\left \{ {{x+y=7} \atop {x-y=-1}} \right.

Since the coordinates of x are the same and both are positive, we can cancel x by subtracting:
\left \{ {{x+y=7} \atop {-(x-y=-1)}} \right. \\ \\2y=8

Divide both sides by 2:
2y/2 = 8/2
y=4

Substitute this into the first equation
x+y=7
x+4=7

Subtract 4 from each side:
x+4-4=7-4
x=3

<u>The coordinates (3, 4) represent the solution to the fourth system.</u>

For the <u>fifth system</u>:
\left \{ {{y=3x-7} \atop {y=2x-5}} \right.

Since y is isolated in each equation, we can set them equation to each other:
3x-7=2x-5

Subtract 2x from each side:
3x-7-2x=2x-5-2x
x-7=-5

Add 7 to each side:
x-7+7=-5+7
x=2

Substitute this into the first equation:
y=3x-7
y=3(2)-7
y=6-7
y=-1

<span><u>The coordinates (2, -1) represent the solution to the fifth system.</u></span> </span>
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Express 4x-1= 3 as a linear equation in two variables​
olga nikolaevna [1]

Hello there!

~I hope you and your family are staying safe and healthy!

4x - 1 = 3

We need to start by saying that y = 3

4x - 1 = y or 4x - y = 1

Thus, the equation above is in the two variable form where y = 3

Thus, the required equation is 4x - y = 1

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The area of a rectangle is 42 ft squared, and the length of the rectangle is 5 ft more than twice the width. Find the dimensions
stira [4]

Answer:

<h3>Length = 12 ft</h3>

Width = \frac{7}{2} ft

Step-by-step explanation:

Given,

Area of rectangle = 42 \:  {ft}^{2}

Width = X

Length = 2x + 5

Now,

x(2x + 5) = 42

2 {x}^{2}  + 5x   = 42

2 {x}^{2}  + 5x - 42 = 0

2 {x}^{2}  + 12x - 7x - 42 = 0

2x(x + 6) - 7(x + 6) = 0

(2x  - 7)(x + 6) = 0

Either

2x - 7 = 0

2x = 0 + 7

2x = 7

x =  \frac{7}{2}

Or,

x + 6 = 0

x = 0 - 6

x =  - 6

Negative value can't be taken.

So, width = \frac{7}{2} ft

Again,

Finding the value of length,

Length = 2x + 5

2 \times  \frac{7}{2}  + 5

7 + 5

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Length = 12 ft

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X+y=3 x-y=1 find the Find the determinants
Mkey [24]

Answer:

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  • Dy = -2

Step-by-step explanation:

<u>System Determinant</u>

The determinant of the system is the determinant of the matrix of coefficients:

  D=\left|\begin{array}{cc}1&1\\1&-1\end{array}\right|=(1)(-1)-(1)(1)=-2

<u>X Determinant</u>

This is the determinant of the matrix with the x-coefficients replaced by the constants.

  D_x=\left|\begin{array}{cc}3&1\\1&-1\end{array}\right|=(3)(-1)-(1)(1)=-4

<u>Y Determinant</u>

This is the determinant of the matrix with the y-coefficients replaced by the constants.

  D_y=\left|\begin{array}{cc}1&3\\1&1\end{array}\right|=(1)(1)-(3)(1)=-2

_____

This problem didn't ask, but the solution is ...

  x = Dx/D = -4/-2 = 2

  y = Dy/D = -2/-2 = 1

  (x, y) = (2, 1)

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