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zepelin [54]
3 years ago
12

Broooo plzzz helpppp

Mathematics
1 answer:
Sindrei [870]3 years ago
5 0

Answer:

3 3/5 + 1 1/4 = <u>4 17/20</u>

7/8 + 1/12 =<u> 23/24</u>

8/9 - 2/5 = <u>22/45</u>

5 1/4 - 2 2/3 = <u>2 7/12</u>

Step-by-step explanation:

3 3/5 + 1 1/4

First convert into improper fractions

= 18/5 + 5/4

Now find the common denominator

72/20 + 25/20 =

97/20

Now simplify

4 17/20

<u>Next problem:</u>

7/8 + 1/12

find the common denominator

21/24 + 2/24 =

23/24

<u>Next problem:</u>

8/9 - 2/5

find the common denominator

40/45 - 18/45 =

22/45

<u>Next problem:</u>

5 1/4 - 2 2/3

First convert into improper fractions

21/4 - 8/3

find the common denominator

63/12 - 32/12 =

31/12

Now simplify

2 7/12

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Helppppppppppp:)))))))))
Whitepunk [10]

Hi there!

We are given the set of ordered pairs below:

\large \boxed{(3, - 1),(2, - 2),(0,2),(2,1)}

1. What is the domain?

  • Domain is a set of all x-values in one set of ordered pairs. So what are the x-values that I am talking about? In ordered pairs, we define x and y which both have relation to each others which we can write as (x,y). That's right, the domain is set of all x-values from ordered pairs.

Therefore, we gather only x-values from (x,y). Hence, the domain is {3,2,0,2}. Whoops! Something is not right. As we learn in Set Theory that we don't write the same or repetitive in a set. Hence, <u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>c</u><u>t</u><u>u</u><u>a</u><u>l</u><u> </u><u>d</u><u>o</u><u>m</u><u>a</u><u>i</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>{</u><u>0</u><u>,</u><u>2</u><u>,</u><u>3</u><u>}</u>

2. What is the range?

  • Because domain is set of all x-values. Then what do you think the range is? That's right! The range is <u>s</u><u>e</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>l</u><u>l</u><u> </u><u>y</u><u>-</u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u>s</u><u>.</u> If you got this right before looking up the underlined words then a handclap for you! So how do we find range? Simple, we just do like finding the domain in the Q1, except we gather the y-values in (x,y) instead and make sure that we don't write same number!

Therefore, gather y-values from the ordered pairs. Hence, <u>t</u><u>h</u><u>e</u><u> </u><u>r</u><u>a</u><u>n</u><u>g</u><u>e</u><u> </u><u>i</u><u>s</u><u> </u><u>{</u><u>-</u><u>2</u><u>,</u><u>-</u><u>1</u><u>,</u><u>1</u><u>,</u><u>2</u><u>}</u>

3. Is the relation a function?

  • All functions are relations but not all relations are functions. Function is a set of ordered pairs where <u>d</u><u>o</u><u>m</u><u>a</u><u>i</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>n</u><u>o</u><u>t</u><u> </u><u>r</u><u>e</u><u>p</u><u>e</u><u>t</u><u>i</u><u>t</u><u>i</u><u>v</u><u>e</u><u> </u><u>o</u><u>r</u><u> </u><u>i</u><u>n</u><u> </u><u>a</u><u> </u><u>s</u><u>e</u><u>t</u><u>,</u><u> </u><u>t</u><u>h</u><u>e</u><u>r</u><u>e</u><u> </u><u>c</u><u>a</u><u>n</u><u>n</u><u>o</u><u>t</u><u> </u><u>b</u><u>e</u><u> </u><u>m</u><u>o</u><u>r</u><u>e</u><u> </u><u>t</u><u>h</u><u>a</u><u>n</u><u> </u><u>o</u><u>n</u><u>e</u><u> </u><u>s</u><u>a</u><u>m</u><u>e</u><u> </u><u>v</u><u>a</u><u>l</u><u>u</u><u>e</u><u>.</u> Consider the following relation: (1,1),(1,2) - Oh, looks like in a set of ordered pairs, there are two same domains which make it only a relation, and not a function. On the other hand, (1,1),(2,2) - Looking good! No same or repetitive domain, making it indeed a function.

Consider the domain from Q1 and see if there are two same values of x in a set. Looks like the relation is not a function since there are same x-values which are 2 in a set, making it only a relation. Hence, the relation is not a function.

These are all 3 answers along with an explanation. Let me know if you have any doubts regarding Relations and Functions.

<em>F</em><em>r</em><em>o</em><em>m</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>Q</em><em>1</em><em>'</em><em>s</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>,</em><em> </em><em>t</em><em>h</em><em>e</em><em>r</em><em>e</em><em> </em><em>a</em><em>r</em><em>e</em><em> </em><em>t</em><em>w</em><em>o</em><em> </em><em>b</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>e</em><em>x</em><em>t</em><em>s</em><em>,</em><em> </em><em>p</em><em>l</em><em>e</em><em>a</em><em>s</em><em>e</em><em> </em><em>c</em><em>h</em><em>o</em><em>o</em><em>s</em><em>e</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em> </em><em>b</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>e</em><em>x</em><em>t</em><em> </em><em>t</em><em>o</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>(</em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>u</em><em>n</em><em>d</em><em>e</em><em>r</em><em>l</em><em>i</em><em>n</em><em>e</em><em>)</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>n</em><em>o</em><em>t</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>f</em><em>i</em><em>r</em><em>s</em><em>t</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>(</em><em>t</em><em>h</em><em>e</em><em> </em><em>o</em><em>n</em><em>e</em><em> </em><em>w</em><em>i</em><em>t</em><em>h</em><em> </em><em>s</em><em>a</em><em>m</em><em>e</em><em> </em><em>2</em><em>'</em><em>s</em><em>)</em><em>.</em><em> </em>

Good luck on your assignment, have a nice day!

4 0
3 years ago
How do you do this question?
ElenaW [278]

x*y' + y = 8x

y' + y/x = 8 .... divide everything by x

dy/dx + y/x = 8

dy/dx + (1/x)*y = 8

We have something in the form

y' + P(x)*y = Q(x)

which is a first order ODE

The integrating factor is u(x) = e^{\int P(x)dx} = e^{\int (1/x) dx} = e^{\ln(x)} = x

Multiply both sides by the integrating factor (x) and we get the following:

dy/dx + (1/x)*y = 8

x*dy/dx + x*(1/x)*y = x*8

x*dy/dx + y = 8x

y + x*dy/dx = 8x

Note the left hand side is the result of using the product rule on xy. We technically didn't need the integrating factor since we already had the original equation in this format, but I wanted to use it anyway (since other ODE problems may not be as simple).

Since (xy)' turns into y + x*dy/dx, and vice versa, this means

y + x*dy/dx = 8x turns into (xy)' = 8x

Integrating both sides with respect to x leads to

xy = 4x^2 + C

y = (4x^2 + C)/x

y = (4x^2)/x + C/x

y = 4x + Cx^(-1)

where C is a constant. In this case, C = -5 leads to a solution

y = 4x - 5x^(-1)

you can check this answer by deriving both sides with respect to x

dy/dx = 4 + 5x^(-2)

Then plugging this along with y = 4x - 5x^(-1) into the ODE given, and you should find it satisfies that equation.

6 0
3 years ago
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s344n2d4d5 [400]
Hope this is correct and hope it helps x

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3 years ago
A rectangle has perimeter 20m. Express the area of the rectangle as a function of the length of one of its sides.
MArishka [77]
<h2>Area of rectangle is 10L-L²</h2>

Step-by-step explanation:

Let L be the length and W be the width.

Perimeter of rectangle = 2 ( L + W )

That is

           2 ( L + W ) = 20

               L + W = 10

                     W = 10 - L

We need to express the area of the rectangle as a function of the length of one of its sides.

           Area = Length x Width

              A = LW

               A = L(10 - L) = 10L-L²

Area of rectangle is 10L-L²

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Answer:

the interest is 195dollars

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