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emmasim [6.3K]
2 years ago
10

Please help I still do not understand these questions

Mathematics
1 answer:
solniwko [45]2 years ago
4 0

Answer:

m = 2/5

Step-by-step explanation:

easy

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Which variable expression represents the word phrase?
slavikrds [6]
I think the answer is A sorry if wrong
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3 years ago
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What is the rate of change and initial value for the linear relation that includes the points shown in the table?
Wittaler [7]

Answer:

Initial value 12, rate of change: -2

Step-by-step explanation:

Step 1

Find the rate of change

we know that

In a linear equation the rate of change is equal to the slope

The formula to calculate the slope between two points is equal to  

m=\frac{y2-y1}{x2-x1}  

we have

A(1,10)\ B(2,8)  

Substitute the values  

m=\frac{8-10}{2-1}

m=-2  ----> rate of change

Step 2

Find the initial value

The equation of  the line into slope-point form is equal to

y-y1=m(x-x1)

Substitute

y-10=-2(x-1)

y=-2x+2+10y=-2x+12

The initial value is the value of y when the value of x is equal to zero (is the y-intercept)

y=-2*0+12=12----> initial value

7 0
3 years ago
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N is m% of what please please please help
Tpy6a [65]

Answer:

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Step-by-step explanation:njjjn

6 0
3 years ago
F is between E and G. EF=3X, EG= 11x-2, FG= 5x+16. Find the value of the variable
Salsk061 [2.6K]

9514 1404 393

Answer:

  x = 6

Step-by-step explanation:

The sum of segments is used:

  EF +FG = EG

  3x +(5x +16) = 11x -2 . . . . substitute given expressions

  8x +18 = 11x . . . . . . . . add 2

  18 = 3x . . . . . . . . subtract 8x

  6 = x . . . . . . divide by 3

The value of the variable is 6.

__

<em>Check</em>

The segment lengths are ...

  EF = 3x = 3·6 = 18

  FG = 5x+16 = 5·6 +16 = 46

  EG = 11x -2 = 11·6 -2 = 64 = 18+46 . . . answer is correct

3 0
3 years ago
Determine the equation of the following polyromial that passes through point (0,6).
Ostrovityanka [42]

Answer:

Cubic polynomial has zeros at x=−1x=−1 and 22, is tangent to x−x−axis at x=−1x=−1, and passes through the point (0,−6)(0,−6).

So cubic polynomial has double zero at x=−1x=−1, and single zero at x=2x=2

f(x)=a(x+1)2(x−2)f(x)=a(x+1)2(x−2)

f(0)=−6f(0)=−6

a(1)(−2)=−6a(1)(−2)=−6

a=3a=3

f(x)=3(x+1)2(x−2)f(x)=3(x+1)2(x−2)

f(x)=3x3−9x−6

3 0
3 years ago
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