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natulia [17]
3 years ago
14

Someone please help?!

Mathematics
1 answer:
natima [27]3 years ago
4 0
May I ask what grade is this?
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Question 2 please answer
8090 [49]

Answer:

The answer is 8

Step-by-step explanation:

if you need proof do to desmos and type it in

6 0
3 years ago
Read 2 more answers
M(x)=x^2-3x<br> N(x)=x-5<br><br> MN(-1)= ?<br><br> A.18<br> B.-20<br> C.-22<br> D. -24
schepotkina [342]

Answer:

d. -24

Step-by-step explanation:

M(-1)=(-1)^2-3(-1)

M(-1)=1+3

M(-1)=4

N(-1)=-1-5

N(-1)=-6

MN(-1)=(4)(-6)

MN(-1)=-24

3 0
4 years ago
A local college has a current total student body enrollment of 12,400. The enrollment is decreasing at a rate of 5% per year. Wh
vovangra [49]

Answer:

FV= 9,716

Step-by-step explanation:

Giving the following information:

Present value (PV)= 12,400

Decrease rate (d)= 5%

Number of periods (n)= 5 years

<u>To calculate the enrolment value (future value) after 5 years, we need to use the following formula:</u>

FV= PV / (1 + d)^n

FV= 12,400 / (1.05^5)

FV= 9,716

8 0
3 years ago
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>
7 0
3 years ago
Will give 25 points! Please Explain! Im so confused! Big test on friday!
Evgen [1.6K]

Answer:

a. y = -x + 7

Step-by-step explanation:

The line with equation y = x - 5 has slope = 1.

A perpendicular to a line has the opposite reciprocal slope of the line, so a perpendicular to the given line has slope -1.

y = -x + b

To find the value of b, we plug in the given point for x and y and solve for b.

y = -x + b

2 = -5 + b

b = 7

y = -x + 7

Answer: a. y = -x + 7

4 0
3 years ago
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