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Anuta_ua [19.1K]
3 years ago
11

Find the equation of the linear function represented by the table below in slope-intercept form. x y 1 2 2 7 3 12 4 17

Mathematics
1 answer:
Mila [183]3 years ago
7 0

Answer:

Step-by-step explanation:

(1, 2) (2, 7)

(7-2)/(2 - 1) = 5/1= 5

y - 2 = 5(x - 1)

y - 2 = 5x - 5

y = 5x - 3

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Are integers irrational
AVprozaik [17]
Well, some integers can be irrational, but not all! :) 
3 0
3 years ago
Read 2 more answers
The mark up on an item is $3.75. If the mark up is 35%, what was the original price? Round
oksian1 [2.3K]

Answer:

$10.71

Step-by-step explanation:

\frac{3.75}{y} =\frac{35}{100}

y × 35 = 3.75 × 100

35y = 375

35y ÷ 35 = 375 ÷ 35

y=10\frac{5}{7}

y = 10.7142857143

10.7142857143 round to 10.71

5 0
3 years ago
What’s the answer to #12? and why
Elena L [17]
Remark
If there are 5 distinct zeros that means either that the x axis is crossed the x axis 5 different places or touched the x axis in 1 place out the 5. Touching in one place means that an even number of roots are the same. 

So let's go through all of them to get an answer of 5.

A has 4 x intercepts. It is not the right answer. We need 5.
B has 4 x intercepts. It is not the right answer. We need 5.
C has 6 x intercepts. Not the one we want.
D has 5 x distinct zeros. The wording is a bit tricky. It does not matter than one of them just touches the x axis. There could be an even number of distinct zeros there, but it only counts as one root.

An example of such a graph is f(x)=\left(\frac{1}{10}(x+2)(x+1.5)(x+1)(x-2)^4(x-3)\right)
 
Answer D <<<<<


6 0
4 years ago
Read 2 more answers
rotation of 90 degrees counterclockwise about the origin, point O, then a reflection across the x-axis reflection across the y-a
SpyIntel [72]

Answer:

The point O can be (x,y)  or (-x,y) or (-x,-y) or ( x,-y) reason being that you haven't given point O lies in which quadrant.

Rotation through 90° counter clockwise

(x,y) = (-y,x)

(-x,y)=(-y,-x)

(-x,-y)=(y,-x)

(x,-y) =(y,x)

Then Reflection across X axis has taken place.

(-y,x) = (-y,-x)

(-y,-x)=(-y,x)

(y,-x)=(y,x)

(y,x)=(y,-x)

Then a reflection across the y-axis has taken place.

(-y,-x)=(y,-x)

(-y,x) = (y,x)

(y,x) =(-y,x)

(y,-x)=(-y,-x)

Then a translation a units to the right and b units up has taken place.

(y,-x)=(y+a,-x+b)

(y,x) =(y+a, x+b)

(-y,x) =(-y+a,x+b)

(-y,-x)=(-y+a,-x+b)

Then a rotation of 180 degrees counterclockwise about the origin has taken place.

(y+a,-x+b)=[-(y+a),-(-x+b)]

(y+a, x+b)=[- (y+a), -(x+b)]

(-y+a,x+b)=[-(-y+a),-(x+b)]

(-y+a,-x+b) =[-(-y+a),-(-x+b)]

Now Again a reflection across the Y axis has taken place.

[-(y+a),-(-x+b)]=[(y+a),-(-x+b)]

[-(y+a),-(x+b)]=[(y+a),-(-x+b)]

[-(-y+a),-(x+b)]=[(-y+a),-(x+b)]

[-(-y+a),-(-x+b)]=[(-y+a),-(-x+b)]

Totally depends on value of a and b on which quadrant these point lies.



7 0
3 years ago
Find the Value of a and b i dare u​
fenix001 [56]

Well it's a dare so I guess I'll do it.

3 0
3 years ago
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