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3 years ago

How do I find the part of a whole of 160% of 19

1 answer:

7 0

$multiply\ percent\ times\ a\ number\\\\
160\%*19=\frac{160}{100}*19=1,6*19=30,4\\\\
160\%\ of\ 19\ is\ equal\ to\ 30,4.$

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If sin(x) = cos(y) for acute angles x and y, how are the angles related?

Vesnalui [34]

Answer:

b. complementary

Step-by-step explanation:

-Complementary angles are angles that add up to 90°.

-These are usually the two acute angles in the right triangle.

#To verify, lets take the two angles 30° and 60°:

$Cos \ 60\textdegree=0.5\\\\Sin \ 30\textdegree=0.5\\\\\therefore Sin \ 30\textdegree=Cos \ 60 \textdegree=0.5$

#We can reverse as:

$Sin \ 60\textdegree=0.86603\\\\Cos \ 30\textdegree=0.86603\\\\\therefore Sin \ 60\textdegree=Cos \ 30\textdegree=0.86603$

Hence, two angles are said to be complimentary if they sum up to 90°.

6 0

3 years ago

I don’t know how to do this

stich3 [128]

To check for continuity at the edges of each piece, you need to consider the limit as $x$ approaches the edges. For example,

$g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}$

has two pieces, $2x+5$ and $x^2-10$ , both of which are continuous by themselves on the provided intervals. In order for $g$ to be continuous everywhere, we need to have

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)$

By definition of $g$ , we have $g(-3)=2(-3)+5=-1$ , and the limits are

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1$

$\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1$

The limits match, so $g$ is continuous.

For the others: Each of the individual pieces of $f,h$ are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.

4 0

3 years ago

Four out of 18 male student and

prisoha [69]

Answer:

Male 2:9

Female 1:7

Step-by-step explanation:

4 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.

5 0

3 years ago

If US=52, ST=30, UT=40, VW=27, and XW=36, find the perimeter of ΔVWX. Round your answer to the nearest tenth if necessary. Figur

Ierofanga [76]

Answer:

The perimeter is 109.8

Step-by-step explanation:

As we can see here, the angles of the triangles are congruent so we can say that both triangles are congruent

When two triangles are congruent, the ratio of their corresponding sides are equal

So we can get the scale factor here by looking at the sides facing each specific angle in both triangles

In the bigger triangle,

The side facing 34 degrees has a length of 30

In the smaller triangle,

The side facing 34 degree has a length of 27

So the scale factor to get the smaller from the bigger is 27/30 = 9/10 or 0.9

Likewise the side facing 51 degrees in the bigger is 40 while for the smaller, it is 36

So the ratio still stands at 36/40 = 9/10 or 0.9

In essence, the smaller triangle will have a perimeter that is 0.9 times that of the bigger

The perimeter of the bigger is simply the sum of the side lengths

We have this as;

(52 + 30 + 40) = 122

so that of the smaller would be;

122 * 0.9 = 109.8

3 0

3 years ago