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malfutka [58]
3 years ago
9

Which equation represents a linear function? y(x – 1) = 9 y – 5 = x(–x + 2) y(y – 1) = x + 25 y – 5 = x – 20

Mathematics
2 answers:
algol133 years ago
7 0

The linear function: y = mx + b.

y(x - 1) = 9 → xy - y = 9   NOT (xy)

y - 5 = x(-x + 2) → y - 5 = -x² + 2x   NOT (x²)

y(y - 1) = x + 25 → y² - y = x + 25   NOT  (y²)

y - 5 = x - 20 → y = x - 15   YES

Gnom [1K]3 years ago
3 0

Answer:

option D is linear

Step-by-step explanation:

y(x - 1) = 9

divide both sides by x-1

y=\frac{9}{x-1} . It represents a rational function

y - 5 = x(-x + 2)

Multiply x inside the parenthesis

y - 5 =-x^2 + 2x

add 5 on both sides

y=-x^2 + 2x+5. It is a quadratic function because we have x^2.

y(y - 1) = x + 25

multiply y inside the parenthesis.

y^2-y = x + 25.  It is a quadratic function because it has y^2

y - 5 = x - 20

add 5 on both sides to get y alone

y= x-15

The equation is in the form of y=mx +b . so it is Linear

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Which equations pass through the points (1, 3) and (7 9)?
ddd [48]

Answer:

f(x)=x+2 Passes through

x-y=2 DOES NOT pass through

Step-by-step explanation:

Method- plug in x and see if the y matches the points. (1,3) (7,9)

f(x)=x+2

f(1)=1+2=3

f(7)=7+2=9

Matches!

x-y=2

1-y=2 --> y=-1

Does not match

8 0
1 year ago
If the sum of $1! + 2! + 3! + \cdots + 49! + 50!$ is divided by $15$, what is the remainder?
kompoz [17]

Answer:

  3

Step-by-step explanation:

All factorials 5 and above are evenly divisible by 15, so have no remainder. Thus, you are interested in ...

  mod(1! +2! +3! +4!, 15) = mod(1 +2 +6 +24, 15)

  = mod(33, 15) = 3

The remainder is 3.

3 0
4 years ago
3x^2+5x+25, when x=3
Vaselesa [24]

[ Answer ]

\boxed{121}

[ Explanation ]

3x^{2} + 5x + 25

X = 3

Replace 3 For X

3*3^{2} + (5 * 3) + 25

Solve

5 * 3 = 15

3 * 3 = 9

9 * 9 = 81

Insert Numbers

81 + 15 + 25

= 121

\boxed{[ \ Eclipsed \ ]}

8 0
4 years ago
Read 2 more answers
Proving circles are similar: Choose any of the transformations that would be a step in proving that Circle A is similar to Circl
Ulleksa [173]

The transformations that would prove that circles A and C are similar are:

  • A. Reflect A over the line y=x
  • C. Dilate A by 3/2

<h3>How to prove that circle A and circle C are similar?</h3>

The circles are given as:

Circle A and B

Assume the following parameters:

  • The center of circle A is (2,3) with a radius of 2
  • The center of circle B is (3,2) with a radius of 3

To start with;

The circle A must be reflected across the line y = x with the following transformation rule:

(x,y) -> (y,x)

So, we have:

(2,3) -> (3,2)

Next, the radius of A must be dilated by 3/2 as follows:

New Radius = 3/2 * 2 = 3

After the transformations, we have the following parameters:

  • The center of circle A is (3,2) with a radius of 3
  • The center of circle B is (3,2) with a radius of 3

Notice that both circles now have the same center and radius.

Hence, both circles are similar

Read more about similar circles at:

brainly.com/question/9177979

7 0
2 years ago
How do you figure out the best buy between 3 items?
hram777 [196]
Lets say u want 3 of them
u multiply each price by 3
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6 0
3 years ago
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