Use the box-and-whisker plots to find the interquartile range for both class B and class A?

Given the function f(x)=-8x2+8x+9,state whether the vertex represents a maximum or minimum point for the function.

Lelechka [254]

8x2+8x+9
=16x+8x+9
=24x+9
Are that right

5 0

2 years ago

1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

-----------------------------------------------------

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

------------------------------

(1) (1) (1)

1 0 1 1

+ 1 1 1

------------------------------

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

2 years ago

Describe and correct the error in finding the tenth term of the arithmetic sequence 4, 12, 20, 28.

ycow [4]

Answer: The tenth term is 76

=======================================================

Explanation:

We use this arithmetic sequence formula to get the nth term

$a_n = a_1 + d(n-1)$

Plug in $a_1 = 4, \ d = 8, \ n = 10$ and you should get the following:

$a_n = a_1 + d(n-1)\\\\a_{10} = 4 + 8(10-1)\\\\a_{10} = 4 + 8(9)\\\\a_{10} = \textbf{76}\\\\$

The tenth term is <u>76</u>

---------

We can verify this by listing out the terms one by one. Start at 4, add on 8 each time, until you generate the 10th term. A table like the one shown below is a good way to keep track of all the terms.

$\begin{array}{c|c}\boldsymbol{n} & \boldsymbol{a_n}\\1 & 4\\2 & 12\\3 & 20\\4 & 28\\5 & 36\\6 & 44\\7 & 52\\8 & 60\\9 & 68\\10 & \textbf{76}\\\end{array}$

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In short, the error is with the "10" in the expression 4+10(8). The student should have used 9 instead. This is because of the n-1 term in $a_n = a_1 + d(n-1)$ which shifts everything one spot to the left.

7 0

1 year ago

1. Make sure your answers and work is SHOWN please and steps are in order.

JulijaS [17]

Been a while since I've done synthetic division.

1a. Let's assume that's supposed to be an equals sign

p(x) = 2x⁴ - 3x³ - 6x² + 5x + 6

possible rational roots have factors of 6 in the numerator and of 2 in the denominator. We'll only worry about negative numerators.

Factors of six: 1,2,3,6, and we don't forget -1,-2,-3,-6

Factors of 2: 1,2

Possible rational roots:

(dividing by 1:) 1,-1,2,-2,3,-3,6,-6

(dividing by 2:) 1/2, -1/2 (2/2=1 is a duplicate, don't have to repeat it), 3/2, -3/2

Possible rational roots: 1,-1,2,-2,3,-3,6,-6, 1/2, -1/2, 3/2, -3/2

Synthetic division, trying x=1,

1 | 2 -3 -6 5 6

2 -1 -7 -2

2 -1 -7 -2 4

Got a remainder of 4, so 1 isn't a root;

Trying x=-1

-1 | 2 -3 -6 5 6

-2 5 1 -6

2 -5 -1 6 0

Zero remainder, found a root, x=-1. This division says

(2x⁴ - 3x³ - 6x² + 5x + 6) / (x + 1) = 2x³ - 5x² - x + 6

Same set of rational roots on the cubic, we continue with x=2

2 | 2 -5 -1 6

4 -2 -6

2 -1 -3 0

Another zero remainder, x=2 is a root. We're left with 2x² - x - 3 = 0, which factors as

(2x - 3)(x + 1) = 0

That's a second factor of x+1. Our final factorization is

2x⁴ - 3x³ - 6x² + 5x + 6 = (x + 1)²(x-2)(2x - 3)

Fourth degree with a positive leading coefficient so goes to +infinity at both ends. Double zero at x=-1, so it's tangent there, just touching the x axis, then down through x=3/2 and up through x=2.

I'll leave the actually sketching and the other two polynomials to you -- that took some time.

4 0

2 years ago

The long side of a box is 6 inches. That is 3 times the length of the short side. How long is the short side? PLEASE HELP NEED A

Montano1993 [528]

Answer:

2 inches

Step-by-step explanation:

If the long side is 6 inches and it's 3x the length of the short side, then you need to divide the long side into 3 parts because the short side is 1/3 the length of the long side. This means that the short side is 2 inches.

5 0

2 years ago

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