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mixas84 [53]
3 years ago
6

A circular walking path has a diameter of 45 yards. what is the distance around the walking path?

Mathematics
1 answer:
Maslowich3 years ago
5 0
3.14 times the radius squared=6358.5
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Giving 40 Points...need help as soon as possible
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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Answer:

WHAT THE MATH IS THIS!!!

Step-by-step explanation:

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Consider the parent function f(x) = x². Also, consider the function g(x) = -3(x-2)² +3. What is the vertex?
Lynna [10]

Answer:

hello :

the vertex is : (2,3) because : g(2) = 3


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4. What is the scale Factor?
nevsk [136]
I know that the scale factor is 2 and the missing value for x is 30
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What is another way to write the expression: t. {14-5} plz I need help on this FAST
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