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3 years ago

Find the perimeter of LMNO. (Hint: Solve for x and y first.)

Mathematics

2 answers:

Damm [24]3 years ago

7 0

6x-7=2x+9

4x=16

x=4

12=y^2+3

9=y^2

3=y -3=y

Nastasia [14]3 years ago

4 0

Step-by-step explanation:

6x -7 = 2x +9

4 x = 16

X= 4

y2 +3 =12

y2 = 9

y2 =3

two side is 17 ,12

Perimeter =( 17+12 )x 2 = 58

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Initially a wheel rotating about a fixed axis at a constant angular deceleration of 0.5 rad/s 2 has an angular velocity of 0 rad

Zigmanuir [339]

Answer:

$\theta = 5.83\ rad$

Step-by-step explanation:

given,

angular deceleration, α = -0.5 rad/s²

final angular velocity,ω_f = 0 rad/s

angular position, θ = 6.1 rad

angular position at 3.9 s = ?

now, Calculating the initial angular speed

$\omega_f^2 = \omega_i^2 + 2 \alpha \theta$

$0 = \omega_i^2 - 2\times 0.5\times 6.1$

$\omega_i = \sqrt{6.1}$

$\omega_i = 2.47\ rad/s$

now, angular position calculation at t=3.9 s

$\theta = \omega_i t + \dfrac{1}{2}\alpha t^2$

$\theta =2.47\times 3.9 - \dfrac{1}{2}\times 0.5\times 3.9^2$

$\theta = 5.83\ rad$

Hence, the angular position of the wheel after 3.9 s is equal to 5.83 rad.

7 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:

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.