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EastWind [94]
3 years ago
5

Find the angle between the hands of a clock at 5:15 A 60 B 67.5 C 75

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0

Answer:

The answer is B 67.5

Step-by-step explanation:

I just did this problem and got it correct

Viktor [21]3 years ago
4 0

Answer:

Option B is correct.

67.5 degree

Step-by-step explanation:

To find the angle between the hands of a clock.

Given that:

Hands of a clock at 5 : 15.

We know that:

A clock is a circle and it always contains 360 degree.

Since, there are 60 minutes on a clock.

\frac{360^{\circ}}{60 minutes} = 6^{\circ} per minutes

so,  each minute is 6 degree.

The minutes hand on the clock will point at 15 minute,

then, its position on the clock is:

(15) \cdot 6^{\circ} = 90^{\circ}

Also, there are 12 hours on the clock

⇒Each hour is 30 degree.

Now, can calculate where the hour hand at 5:00 clock.

⇒5 \cdot 30 =150^{\circ}

Since, the hours hand is between 5 and 6 and we are looking for 5:15 then :

15 minutes is equal to \frac{1}{4} of an hour

⇒150+\frac{1}{4}(30) = 150+7.5 = 157.5^{\circ}

Then the angle between two hands of clock:

⇒\theta = 150.75 -90 = 67.5^{\circ}

Therefore, the angle between the hands of a clock at 5: 15 is: 67.5 degree.

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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
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