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3 years ago

6

What are the x-intercepts of the graph of the function f(x) = x2 + 5x − 36? (−4, 0) and (9, 0) (4, 0) and (−9, 0) (−3, 0) and (1

2, 0) (3, 0) and (−12, 0)

2 answers:

kupik [55]3 years ago

5 0

Answer:

(4,0) & (-9,0)

Step-by-step explanation:

Just took the test.

MatroZZZ [7]3 years ago

3 0

Answer:

(4,0) & (-9,0)

Step-by-step explanation:

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Can someone plz help me!!

AlladinOne [14]

Answer:

145°

Step-by-step explanation:

upper left angle inside triangle is 180-137, or 43°

lower right angle inside triangle is 180-(43 + 102) or 35°

'?' measures 180 - 35 which is 145°

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3 years ago

Which of these tables represents a linear function?

tangare [24]

one second the think I

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4 years ago

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MatroZZZ [7]

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astra-53 [7]

Answer:

1. 44 + 3x

2. 2y - 8

3. x - 6

15. $5\frac{7}{8}$

16. $6\frac{1}{3}$

17. $3\frac{7}{9}$

Step-by-step explanation:

1. 7² + 2² - 5 - 4 + 3x

49 + 4 - 5 - 4 + 3x

53 - 5 - 4 + 3x

48 - 4 + 3x

44 + 3x

2. - y - 5 + y + 2(2y-y) - 3

-y - 5 + y + 4y - 2y -3

-y - 5 + 5y - 2y - 3

4y - 2y - 5 - 3

2y - 8

3. 5x -3 - x - 3(x + 1²)

5x - 3 - x - 3x - 3

4x - 3x - 3 -3

x - 3 -3

x - 6

15. $7\frac{1}{4} - 1\frac{3}{8}$

= $7 \frac{2}{8} - 1\frac{3}{8}$

= $\frac{58}{8} - \frac{11}{8}$

= $\frac{47}{8}$ → $5\frac{7}{8}$

16. $9 - 2\frac{2}{3}$

= $\frac{54}{6} - 2\frac{4}{6}$

= $\frac{54}{6} - \frac{16}{6}$

= $\frac{38}{6}$ → $6\frac{2}{6}$ → $6\frac{1}{3}$

17. $10\frac{1}{3} - 6\frac{5}{9}$

= $10 \frac{3}{9} - 6\frac{5}{9}$

= $\frac{93}{9} - \frac{59}{9}$

= $\frac{34}{9}$ → $3\frac{7}{9}$

Hope this helps.

6 0

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