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Lerok [7]
3 years ago
11

Helppppppppppppppppppppppppp

Mathematics
2 answers:
aev [14]3 years ago
6 0
We evaluate the function at a known point.
 We have that for x = 0
 y = sin (x + 90)
 y = sin (0 + 90)
 y = sen (90)
 y = 1
 Then, the function sought will be:
 y = sin (x + 90)
 Answer:
 y = sin (x + 90)
 option A
Firdavs [7]3 years ago
4 0
The allknowing wanser is c

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How to find the median
KengaRu [80]

<em>We </em><em>can </em><em>find </em><em>the </em><em>median </em><em>by </em><em>using </em><em>the </em><em>formula:</em>

<em>(</em><em>N+</em><em>1</em><em>/</em><em>2</em><em>)</em><em> </em><em>th </em><em>item</em>

<em>For </em><em>example:</em>

<em>Given </em><em>data:</em><em> </em><em>4</em><em>,</em><em>6</em><em>,</em><em>3</em><em>,</em><em>5</em><em>,</em><em>7</em>

<em>We </em><em>have </em><em>to </em><em>arrange </em><em>the </em><em>data </em><em>into </em><em>ascending</em><em> </em><em>order</em><em>:</em>

<em>=</em><em>3</em><em>,</em><em>4</em><em>,</em><em>5</em><em>,</em><em>6</em><em>,</em><em>7</em>

<em>N(</em><em>total </em><em>no </em><em>of </em><em>items)</em><em>=</em><em>5</em>

<em>Now,</em>

<em>Median=</em><em>(</em><em>N+</em><em>1</em><em>/</em><em>2</em><em>)</em><em>th </em><em>item</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>(</em><em>5</em><em>+</em><em>1</em><em>/</em><em>2</em><em>)</em><em> </em><em>th </em><em>item</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>(</em><em>6</em><em>/</em><em>2</em><em>)</em><em> </em><em>th </em><em>item</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>3</em><em>r</em><em>d</em><em> </em><em>item</em>

<em>Median=</em><em>5</em>

<em>Hope </em><em>it</em><em> </em><em>helps</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

3 0
2 years ago
Read 2 more answers
Is 4/5 greater or less than 45/100 how do you know
rusak2 [61]

Answer:

4/5 is grater than 45/100

Step-by-step explanation:

Why?

because in percentage you make the bottom # a 100 one of them already has that and when it has 100 on the bottom it tells you that the top # is the percentage so 45/100 is 45%

now to make 4/5 a percentage what times 5 = 100?

20 does

so you times the top and bottom by 20

4*20= 80

now your fraction is 80/100  (top # tells you the %)

it's 80%

80% is grater than 45%

5 0
2 years ago
Read 2 more answers
Point B lies between points A and C, and all three points lie on point AC, which of the following is not true? A. Point B lies o
loris [4]

Answer:

C. Point A lies on ray BC

Step-by-step explanation:

Points A and C can be connected by a segment which would be a measure of the distance between the points. Locating point B between AC, makes the three points lying on segment AC.

A ray extends from a point to infinity, a line extend to infinity on both sides, while a segment is known to have two endpoints. Therefore, points AC are the end points of the segment AC, and point B between this segment confirms that point B lies on the segment AC. Therefore, Point A lies on ray BC is not correct.

8 0
3 years ago
I am struggling with this problem
irinina [24]

Answer:

b

Step-by-step explanation:

8 0
3 years ago
Let P (x, y) be the statement "Student x has taken class y," where the domain for x consists of all students in your class and f
Rufina [12.5K]

Answer: Hello mate!

we know that p(x,y) means "Student x has taken class y"

and the used symbols are:  

∃: this means "existence", you use this symbol to say that there exists at least one object that makes true the sentence.

∀: this means "for all", you use this symbol to say that the sentence is true for all the elements, then:

a) ∃x∃yP (x, y)

"exist at least one student x, that took at least one class y"

b) ∃x∀yP (x, y)

"exist at least one student x, that took all the classes y"

c) ∀x∃yP (x, y)

"every student x, took at least one class y"

d) ∃y∀xP (x, y)

"exist at least one class y, that has been taken by all the students x"

e) ∀y∃xP (x, y)

"for every class y, there is at least one student x that took the class"

f) ∀x∀yP (x, y)

"all the students x took all the classes y"

4 0
3 years ago
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