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2 years ago

What is the average rate of change for this quadratic function for the interval

Mathematics

1 answer:

Romashka [77]2 years ago

7 0

Answer:

x 2 − 8 x + 15

Step-by-step explanation:

( x − 3 ) ( x− 5)

Expand ( x − 3 ) ( x − 5 )

using the FOIL Method.

x ⋅ x + x ⋅ − 5 − 3 x − 3 ⋅ − 5

Simplify and combine like terms.

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Each marble bag sold by Lashonda's Marble Company contains 5 green marbles for every 3 red marbles. If a bag has 24 red marbles,

lubasha [3.4K]

Hope this helps but the answer is that if there are 24 red marbles in a bag then there are 40 green marbles.

4 0

3 years ago

If the input, x, is 16, which equation results when output of 4?

Bad White [126]

Answer:

d) y = x-12

Step-by-step explanation:

x is the input

y is the output

d) 4 = 16 - 12

7 0

3 years ago

How to find the perimeter of the triangle

Sliva [168]

Answer:

A. 26.2

Step-by-step explanation:

To find the perimeter of the triangle, you have to find the distances of all three lines and add them up.

Line AB

Let's start off by finding the distance of line AB.

We will use the formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point B is (4,-3).

To substitute the values, it will get to $d = \sqrt{(4--2)^{2}+(-3-5)^{2}}$ which in other words is $d = \sqrt{(4+2)^{2}+(-3-5)^{2}}$ .

Now we have to solve the parenthesis to get $d = \sqrt{(6)^{2}+(-8)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{36+64}$ .

Now we have to simplify the square root to $d = \sqrt{100}$ . In other words, that is $d = 10$ .

Line AB = 10

Line BC

Now let's find the distance of line BC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point B is (4,-3) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0-4)^{2}+(-6--3)^{2}}$ which in other words is $d = \sqrt{(0-4)^{2}+(-6+3)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(-4)^{2}+(-3)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{16+9}$ .

Now we have to simplify the square root to $d = \sqrt{25}$ . In other words, that is $d = 5$ .

Line BC = 5.

Line AC

Now let's fine the distance of line AC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0--2)^{2}+(-6-5)^{2}}$ which in other words is $d = \sqrt{(0+2)^{2}+(-6-5)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(2)^{2}+(-11)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{4+121}$ .

Now we have to simplify the square root to $d = \sqrt{125}$ . In other words, that is $d = 5\sqrt{5}$ . To round that, $d = 11.2$ .

Line AC = 11.2.

Perimeter of Triangle ABC

Perimeter of Triangle ABC = Line AB + Line BC + Line AC.

Perimeter of Triangle ABC = 10 + 5 + 11.2

Perimeter of Triangle ABC = 26.2

Hope this helped! If not, please let me know <3

3 0

3 years ago

I need help with this pleasee

never [62]

Answer:2 2/15

Step-by-step explanation:

4/5= 12/15

+1 1/3 =5/15

1 and 17/15

17/15= 1 and 2/15 +1= 2 and 2/15

8 0

3 years ago

Plzzzzzzzzzzz helppppppp

Svetach [21]

Answer:

1st answer

Step-by-step explanation:

4 0

3 years ago