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3 years ago

Find the area of the following rectangles. 5 7/8 km x 18/4 km

Mathematics

2 answers:

tamaranim1 [39]3 years ago

7 0

<span> is the answer a Polygon</span>

PolarNik [594]3 years ago

5 0

The answer is: 26 437 500 m<span>2</span>

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PLEASE HELP PICTURE SHOWN

Rainbow [258]

Your answer is B. 2
This is because to rationalise the denominator, we need to multiply it by (3 - √7), so we get:
(3 + √7)(3 - √7)
3 × 3 = 9
3 × √7 = 3√7
3 × -√7 = -3√7
√7 × -√7 = -7
So all in all you get 9 - 7 which is 2.

I hope this helps!

4 0

3 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

Help no explanation just answers plz, theres no multi choice

maxonik [38]

Answer:

A b or c

Step-by-step explanation:

Easy shahhshhshs free points hehe

4 0

3 years ago

8c^3 d^2 divided 4cd^2

Lana71 [14]

Answer: 2c^2

Step by step explanation:

4 0

3 years ago

Given the functions f(x) = 2x + 5 and g(x) = x2 + 8, which of the following functions represents f(g(x)] correctly?

lakkis [162]

Answer:

Choice 4.

Step-by-step explanation:

f(g(x))

Replace g(x) with x^2+8 since g(x)=x^2+8.

f(g(x))

f(x^2+8)

Replace old input,x, in f with new input, (x^2+8).

f(g(x))

f(x^2+8)

2(x^2+8)+5

Distribute:

f(g(x))

f(x^2+8)

2(x^2+8)+5

2x^2+16+5

Combine like terms:

f(g(x))

f(x^2+8)

2(x^2+8)+5

2x^2+16+5

2x^2+21

7 0

3 years ago