–215, 214, 2.23,

If g (x) = 3x+7, then which of the following is the solution to g (x) = 1

BabaBlast [244]

Answer:

the answer might be 10 not really sure

4 0

3 years ago

What is the area of the yellow region? INEED HELP PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE!!!

True [87]

Answer:

A=¼πr²

=¼(3.142×5²)

=19.64 square miles

8 0

3 years ago

Blain is two years older than three times Jillians age. Jillian is also 16 years younger than Blain. How old is Blain?

Ipatiy [6.2K]

The age of Blain is 23 years old

<h3><u>Solution:</u></h3>

Let the age of Blain be "a" and age of Jillian be "b"

Given that Blain is two years older than three times Jillians age

So we can frame a equation as:

age of blain = 2 + 3(age of Jillian)

a = 2 + 3b ----- eqn 1

Also given that Jillian is also 16 years younger than Blain

Age of Jillain = Age of Blain - 16

b = a - 16 ---- eqn 2

Substitute eqn 2 in eqn 1

a = 2 + 3(a - 16)

a = 2 + 3a - 48

a - 3a = -46

-2a = -46

a = 23

Thus the age of Blain is 23 years old

4 0

3 years ago

Need help simple explanation if possible

photoshop1234 [79]

Answer:

m < A = 60º

m < B = 30º

Step-by-step explanation:

The given sides on this triangle are: 6, $6\sqrt{3}$ , and 12

Any triangle with the angles of 30º - 60º - 90º always has side lengths in this proportion:

x, $x\sqrt{3}$ , 2x

We can line this up with the given sides. If x is 6, then 2x would be 12.

x : $x\sqrt{3}$ : 2x = 6 : $6\sqrt{3}$ : 12

Angle B is across from 6, the shortest side. That also means that it corresponds to x, or the smallest angle in the proportion, 30º.

m < B = 30º

Solving for < A:

Method 1) Sum of Angles in a Triangle

Since we already know that one angle is right and therefore 90º and m < B is 30º, we can subtract these from the total sum of angle measures in a triangle to get the last angle, < A.

180º - 90º - 30º = 60º

m < A = 60º

Method 2) Using the second part of the proportion

Since m < A is across from the second largest side, we know that it is equal to $x\sqrt{3}$ ( $6\sqrt{3}$ in this question) or 60º in the angle proportion.

This means that m < A = 60º

Let me know if you have any questions!

4 0

2 years ago

Read 2 more answers

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

3 years ago

–215, 214, 2.23, –2=