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Aliun [14]
3 years ago
13

On Friday, Carlotta had 6/8 of her paper left to write. She wrote 2/6 of the paper on Saturday and ¼ of the paper on Sunday. At

the end of the weekend, what fraction of her paper does Carlotta have left to write?
Mathematics
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

\frac{1}{6}

Step-by-step explanation:

<u>Skills needed: Fraction Subtraction, Least Common Denominator (LCD)</u>

1) \frac{6}{8} - \frac{2}{6} - \frac{1}{4} is the operation you would do.

--> You would do this because, Carlotta has \frac{6}{8} of the paper left, and does one fraction amount (\frac{2}{6}) on Saturday & does another fraction amount (\frac{1}{4}) on Sunday.

-->  Using subtraction is the best way

2) Now, we need to figure out the Least Common Denominator for all 3 fractions.

8,6,4 are the denominators.

The lowest number that is a factor of all them is 24.

So, it would then become \frac{18}{24} - \frac{8}{24} - \frac{6}{24}

This would become: \frac{18-8-6}{24} (18-8-6 = 4)

So then: \frac{4}{24} is your answer.

This simplifies into \frac{1}{6}.

Hope this helps and have a nice day!! :) :]

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2(7) = 14

Step-by-step explanation:

When there is a number in the brackets, we want to input that into our function and get the output. In other words, we simply substitute 7 for x

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6 points are placed on the line a , 4 points are placed on the line b . How many triangles is it possible to form such that thei
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Answer:

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Step-by-step explanation:

Using side b as the base, 4 points makes 3 bases (the space in between).  With three bases, you can have 3 bases of 1 segment, 2 bases of 2 segments, and 1 base of 3 segments.  This equals 6 bases. Each of these can connect to a point on line a. 6x6=36

Using side a as the base, 6 points makes 5 bases.  With 5 bases, you can have 5 bases of 1 segment, 4 bases of 2 segments, 3 bases of 3 segments, 2 bases of 4 segments, and 1 base of 5 segments.  This equals 15 bases.  Each of these can connect to a point on line b.  15x4=60

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Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.

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(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

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