The graph of y = x2 - 4x is shown on the grid.

The minimum size of a photo book ordered through a website is 6 pages. The website charges $1.70 per page for the first 6 pages.

m_a_m_a [10]

Well Okay, 1.70 x 6 + .90x, Your pretty much taking 1.70 multiply by 6 then adding .90 for every page after.

6 0

3 years ago

HELP PLEASE <br><br>must show work

matrenka [14]

Answer:

1. 4n^3

2. 4k^7

3. 3

4. -30x

5. -6

Step-by-step explanation:

1. The prime factorization of 12 is 2 x 2 x 3 and the prime factorization of 16 is 2 x 2 x 2 x 2. When you look at these two expressions you can see the common factors of these two numbers are 2 x 2, which is 4. Next, we look at the GCF of the N's which would be n^3 since n^5 has three N's in it. Therefore, we get 4n^3 when we multiply the two together.

2. The factors of 8 are 1, 2, 4, and 8. Out of these, 1, 2, and 4 are the only factors that 20 shares with it and 4 is the greatest. Then, we look at the K's and the GCF of the K's is k^7 since k^8 has seven K's. We multiply the two and we get 4k^7.

3. Since one of the numbers of the three given here does not include the variable n, there will not be any N's in the GCF of the three, so we don't have to worry about that. Now, we just find the GCF of 18, -24, and -21. The factors of 18 are 1, 2, 3, 6, 9, and 18, the factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24, and lastly, the factors of 21 are 1, 3, 7, and 21. From these, 3 is the biggest common divisor, therefore the GCF is 3.

4. Between the two X's, X^1 is the biggest amount of X's this GCF has, so the final GCF will be some constant multiplies with X. Since we are dealing with bigger numbers on this problem, we should use prime factorization. The prime factorization of 90 is 2 x 3 x 3 x 5, and the prime factorization of 120 is 2 x 2 x 2 x 3 x 5. From these expressions, we take the biggest amount of each common factor as we can. Since these expressions both have 2, we take the smaller amount of 2's which is one two. Then we get one three from both expressions, and one five as well. 2 times 3 times 5 equals 30, therefore, we get -30x, and not 30x, because both of these numbers are negatives.

5. All of these numbers do not have an x, so there won't be an x in our GCF. Another method of quickly finding the GCF of numbers is to look at the smallest number's factors first to see what factors it shares with the other numbers. The factors of 12 are 1, 2, 3, 4, 6, and 12. 42 and 30 do not have the factor 12, so we can go down the list and see if 42 and 30 share the factor 6, which they do since 6 times 7 is 42 and 6 times 5 is 30. Since all of these numbers share the negative sign, the GCF of these three numbers is -6.

5 0

3 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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7 0

4 years ago

Consider the following region R and the vector field Bold Upper F.

Natali5045456 [20]

Answer and Step-by-step explanation:

Workings and solutions to a, b, & c can be viewed in the snapshot below:

7 0

4 years ago

The table below shows 10 data values 125 138 132 140 136 136 126 122 135 121 What values of minimum, Q1, median, Q3, and maximum

Troyanec [42]

We have to put the numbers in order..
(121,122,125, 126,132),M,(135,136,136,138,140)

minimum : (smallest number) = 121
Q1 : 125
Q2 (median) : (132 + 135) / 2 = 267/2 = 133.5
Q3 : 136
maximum : (largest number) = 140

4 0

3 years ago

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