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IgorLugansk [536]
2 years ago
7

There are 60 pupils who joined the Math Quiz Competetion. 40 were boys and 20 were girls. During the Awarding ceremony, it was f

ound out that 20% of the boys and 15% of the girls won and received prizes. How many received prizes?
Please Answer In English
No link Kailangan ko na po ngayon!! :)​
Mathematics
1 answer:
Pavlova-9 [17]2 years ago
6 0

Answer:

8 pupils received prices

Step-by-step explanation:

20% of 40 = 8 (8 boys)

15% of 20 =3  (3 girls)

8+3 = 11

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Estimate the answer to the following problem to the nearest 10 13+46+87= a. 130 b.155 c.160 d.150
Lunna [17]
150

13+46= 59
59+87= 146

Round 46 to the nearest ten, which in this case would be 50 because the last number is larger than a 5. 

Then add the 1 back in to make it 150.
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How do animals breakedown rocks.
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A,B

Step-by-step explanation:

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What is the area of this figure
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108

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Two trains on opposite tracks leave the same station at the same
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20 minutes( 0.3333 hours) after they leave the station will they be 50 km apart, if the Two trains on opposite tracks leave the same station at the same time.

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2 years ago
*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts
Nataly [62]

Answer: Assuming the function is g(x)=\frac{2x+1}{x-3}:

The x-intercept is (\frac{-1}{2},0).

The y-intercept is (0,\frac{-1}{3}).

The horizontal asymptote is y=2.

The vertical asymptote is x=3.

Step-by-step explanation:

I'm going to assume the function is: g(x)=\frac{2x+1}{x-3} and not g(x)=2x+\frac{1}{x}-3.

So we are looking at g(x)=\frac{2x+1}{x-3}.

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

0=\frac{2x+1}{x-3}

A fraction is only 0 when it's numerator is 0.  You are really just solving:

0=2x+1

Subtract 1 on both sides:

-1=2x

Divide both sides by 2:

\frac{-1}{2}=x

The x-intercept is (\frac{-1}{2},0).

The y-intercept is when x is 0.

Replace x with 0.

g(0)=\frac{2(0)+1}{0-3}

y=\frac{2(0)+1}{0-3}  

y=\frac{0+1}{-3}

y=\frac{1}{-3}

y=-\frac{1}{3}.

The y-intercept is (0,\frac{-1}{3}).

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is y=\frac{2}{1}.  After simplifying you could just say the horizontal asymptote is y=2.

Or!

I could do some division to make it more clear.  The way I'm going to do this certain division is rewriting the top in terms of (x-3).

y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}

y=2+\frac{7}{x-3}

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0
3 years ago
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