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3 years ago

What are the solution(s) to the quadratic equation 40 − x2 = 0?

Mathematics

2 answers:

alina1380 [7]3 years ago

8 0

Answer:

$x=+2\sqrt{10}$

$x=-2\sqrt{10}$

Step-by-step explanation:

$40-x^2=0$

To solve for x , subtract 40 from both sides

$-40+40-x^2=0-40$

$-x^2=-40$

Divide both sides by -1

$x^2= 40$

Take square root on both sides

$\sqrt{x^2} =\sqrt{40}$

$x=+-2\sqrt{10}$

The values of x are

$x=+2\sqrt{10}$

$x=-2\sqrt{10}$

Anni [7]3 years ago

4 0

X = 20
First, you move the 40 to the other side.
-2x=-40
Then, you divided -40 by -2.
x= -40/-2
x= 20
x is 20 because the negatives cancel out making it positive.

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Answer:

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2 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:

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.

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

2 years ago

Which functions domain consists of all real numbers except -2?

Eddi Din [679]

Answer:

option (a) f(x)= 1/x+2

Step-by-step explanation:

(a) f(x) = 1/ x+2

To find the restriction for domain , we set the denominator =0 and solve for x

x+2 =0, so x=-2

When x=-2 then denominator becomes 0 that is undefined.

So, domain is all real numbers except -2

(b) f(x)= 2x

In this function, there is no denominator or square root or log function

so there is no restriction for x, hence domain is all real numbers

In this function, there is no denominator or square root or log function

so there is no restriction for x, hence domain is all real numbers

f(x) = 1/ sqrt(x+2)

if we have square root in the denominator then we set the denominator >0 and solve for x. because square root of negative values are undefined

x+2>0, x>-2

Hence domain is all real numbers that are greater than -2

8 0

3 years ago

Use the drop-down menus to choose steps in order to correctly solve 4k−6=−2k−16−2 please help just need steps

luda_lava [24]

4k-6=-2k-16-2
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2 years ago