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3 years ago

7

Suppose a plant grow one inch every year if X is how old the plant is (years) and f(x) is how tall the plant is (in inches) what

type of function is f(x)?
A. constant function
B. identity function
C. absolute value function
D. step function

1 answer:

miskamm [114]3 years ago

5 0

It’s C. absolute value function

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Carrie is buying bulk candy for a high school fundraiser. She buys 1000 candy bars for $350. What is the unit cost per candy bar

liubo4ka [24]

1000 candy bars ... $350
1 candy bar ... $x = ?

If you would like to know what is the unit cost per candy bar, you can calculate this using the following steps:

1000 * x = 350 * 1
1000 * x = 350 /1000
x = 350 / 1000
x = $0.35

Result: She can buy 1 candy bar for $0.35, so the correct result would be B) $.35.

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3 years ago

Systems By Graphing - Solve All Correctly For a Thanks, 5-Stars, And A Brianliest!

BartSMP [9]

Answer:

Step-by-step explanation:

Amari= Graph C and Solution is (-6,-2)

Bella= Graph A and Solution is (3,4)

Carl= Graph B and Solution is (0,-3)

0 solutions because the slopes are the same. The lines will never cross because they are at the exact same angle.

No, I do not agree with Joey because the lines have different slopes and will lead to the system cross which is the solution.

1st graph: y=2x-1 and y=7 solution #1= (4,7)

2nd graph: y=-2x-3 and y=1/2x solution #2= (-2,1)

3rd graph: y=x and y= -1/5+6 solution #3= (5,5)

I had this exact same assignment a few months ago, my teacher didn't use the 2nd slide but I had the 1st and 3rd slide so this should help!

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3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

<h3>(a)</h3>

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

<h3>(b)</h3>

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

3 years ago

I need help please.

GuDViN [60]

Answer:

The factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....

Step-by-step explanation:

The given expression is:

2q²-5pq-2q+5p

Make a pair of first two terms and last two terms:

(2q²-5pq) - (2q-5p)

Now factor out the common factor from each group.

Note that there is no common factor in second group. So we will take 1 as a common factor.

q(2q-5p) -1(2q-5p)

Now factor the polynomial by factoring out the G.C.F, 2q-5p

(2q-5p) (q-1)

Thus the factors of 2q²-5pq-2q+5p are (2q-5p) (q-1)....

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3 years ago

Abby tosses a coin off a bridge into the stream below. The distance, in feet, the coin is above the water is modeled by the equa

Harlamova29_29 [7]

Answer: do it yourself

Step-by-step explanation:

5 0

3 years ago