Please lang po paki answer tong modules ko please

Mathematics

1 answer:

Blizzard [7]3 years ago

8 0

Answer:

1) cancel common factors

anything in the numerator in one fraction that equals to a denominator in the other fraction should cancel because they equal one

$\frac{2}{9} \times \frac{9}{2} = 1$

also (10/10) = 1

this also answer the next two Q

4) multiply fractions

multiply what ever in the numerator and multiply in denominator and simplify if requires

(3/10) × ( 5/6) = 15 / 60

3× 5 = 15

10 × 6 = 60

15 / 60 can be simplified by dividing top and bottom by 3

15 ÷ 3 = 5

60 ÷ 3 = 20

= 5/20

this also can be simplified by dividing by 5

5 ÷ 5 = 1

20÷ 5 = 4

= ( 1/4)

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Lulu tells her little brother, Jack, that she is holding 20 coins, all of which are either dimes or quarters. They have a value

RSB [31]

Given :

Lulu tells her little brother, Jack, that she is holding 20 coins, all of which are either dimes or quarters.

They have a value of $4.10. She says she will give him the coins if he can tell her how many of each she is holding.

To Find :

How many dimes and quarters in Lulu holding?

Solution :

4.10 dollar = 410 cents

Let, number of dimes and quarter are x and y.

So, x + y = 20 ...1)

Also, 10x + 25y = 410 ...2)

Putting value of 1) in equation 2), we get :

10x + 25( 20 - x ) = 410

10x + 500 - 25x = 410

15x = 90

x = 6

y = 14

Therefore, number of dimes and quarters are 6 and 14 respectively.

Hence, this is the required solution.

3 0

3 years ago

If y varies directly as x and x = 9 when y = 15 , find y when x = 33

poizon [28]

This would practically be very reliable of course. We can see in the sentence above that y and x, and x and y are both depending on each other. So, now that this has been considered, we can see that whenever "x" would be a certain number, this would show you that "y" would also be a certain number, and depending how much "x" or "y" would move, it would also change these variable no matter how they move.

This would be a matter of division. As we can see above on how 9 and 15 could be divisible down the line as they multiply.

Therefore, we would then see how many times would "x" which is 9, how many times that would go in 33.

Let's start!

$\boxed{33\div9= \ (3.6)}$

We take this "3.6" away from this, and we would plug it into "y".

We then do the following:

$\boxed{3.6*15=55}$

Therefore, when "x" = 33, "y" would then equal 55.

Your answer: $\boxed{\boxed{\bf{y=55}}}$