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melisa1 [442]
3 years ago
8

One side of a right triangle is 2 cm shorter than the hypotenuse and 7 cm longer than the third side. find the lengths of the si

des of the triangle.
Mathematics
1 answer:
astra-53 [7]3 years ago
8 0
The right angles triangle has three sides including the hypotenuse which is opposite to the right angle.
If the hypotenuse - H
Then one side is - H - 2
Third side is 7 cm shorter then H-2
Therefore it’s H -2-7 = H -9
According to Pythagoras theorem
Square of hypotenuse = sum of the squares of the other 2 sides
H² = (H-9)² + (H-2)²
H² = H² -18H + 81 + H² -4H + 4
0 =H² -22H +85
This is a quadratic equation
0 = H² - 17H -5H +85
0 = H (H - 17) -5(H-17)
(H-17) (H-5) =0
H -17=0 or H-5=0
H could be 17 cm or 5 cm
If H =5 cm then
Side 1 -H-2 = 3 cm
And side 2 - H-9 = -4 cm
Since lengths cannot be negative values
H isn’t 5 cm
Therefore H =17 cm
1 side is H-2 = 15 cm
2nd side H -9 = 8 cm
Three sides are 17 cm, 15 cm and 8 cm
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Without calculating, explain how many decimal places are in the product 0.6 X0.4 X 1.9. Rewrite the expression as the product of
Vladimir79 [104]
There's only one decimal point in the answer. This is quite obvious. Regardless of whether or not your adding, subtracting, dividing, or multiplying, there's really only going to be one decimal, ever. This doesn't change, However, what DOES change are decimal PLACES.

Looking at the problem, you see both .6 and .4 are placed to the right of the decimal, otherwise known as the tenths place (1/10). 

Looking at 1.9, you see you have one number (1) placed to the left of the decimal, the ones place, and another number placed to the right of the decimal (9), in the tenths place. 

Well, this is simple. You can look and see that your multiplying fractions against a whole number, which will of course make this whole number even smaller. By how much though??

Well, you can really tell this by looking at how MANY numbers are behind the decimal. You have .9, .4, and .6

That's 3 numbers behind the decimal place. This may not help you solve the problem itself, but this tells you here (ignoring the 1, that will be taken out immediately while solving and turned into a 0) that you will have 3 numbers placed from the tenths place all the way to the thousandths place.

Basically, this tells you that you'll have three decimal places to the right of the decimal. 

As for rewriting your problem to better show this, that's simple too.

All you have to do is simple multiplication, just in the tenths place. How? You know that 6*4 (referring to .6 and .4) is 24, that means that .6*.4 will be .24

How does this support your answer before? .24 has TWO numbers to the right of the decimal. That's TWO decimal places. With the .9 in 1.9, that's still THREE decimal places to the right of the decimal, and your answer would be the same.

So, here's it re-written.
1.9*.24=X

~Hope this helps!


4 0
3 years ago
Yes I wanted to know what is the answer ​
jek_recluse [69]

Answer:

14 flats

5 plants

Step-by-step explanation:

First you would do 89/6 but since that doesn’t divide nicely you can do 90/6 to get 15 then subtract 1 since the original number was 89 and you would get 14

The 14 times 6 would get you 84

Then subtract 84 from 89 and you get 5 plants left over ^^

7 0
2 years ago
En una urna hay 200 esferas, el 25% son rojas, 1/4 son amarillas, la mitad de lo que queda son azules y el resto es verde. Calcu
masha68 [24]

Answer:

I no speak spanish

Step-by-step explanation:

6 0
3 years ago
Urn 1 contains 3 blue tokens and 2 red tokens; urn 2 contains 2 blue tokens and 4 red tokens. All tokens are indistinguishable.
Dmitry_Shevchenko [17]

Answer:

RR = 0.4

RB = 0.3

BB = 0.22

BR = 0.30

Step-by-step explanation:

P( Urn 1 ) = 2/6 = 1/3

P( Urn 2 ) = 1 - 1/3 = 2/3

Urn 1 contains : 3 blue and 2 red

P( blue | urn 1 ) = 3/5 ( with replacement ) , P( blue | urn 1 ) = 3/4 ( without replacement )

P( red | urn 1 ) = 2 / 5 ( with replacement ) , P(red | urn 1 ) = 1/2 ( without replacement )

Urn 2 contains : 2 blue and 4 red

P ( blue | urn 2 ) = 1/3 ( with replacement ) , P( blue | urn 2 ) = 2/5 ( without replacement )

P ( red | urn 2 ) = 2/3 ( with replacement) , P( red | urn 2 ) = 4/5 ( without replacement )

Determine

<u>i) Possible outcomes when two tokens are drawn from either Urn without replacement </u>

RR = [[ ( 2/5 * 1/3 ) + ( 2/3 * 2/3 ) ] * [( 1/2 * 1/3 ) + ( 4/5 * 2/3 ) ]] = 0.4

RB = [[ (2/5 * 1/3 ) + ( 2/3 * 2/3 ) ] * [ ( 3/4 *1/3 ) + ( 2/5 * 2/3 ) ]] ≈ 0.3

BB = [[ ( 3/5 * 1/3 ) + ( 1/3 * 2/3 ) ] * [ ( 3/4 *1/3 ) + ( 2/5 * 2/3 ) ]] ≈ 0.22

BR = [[ ( 3/5 * 1/3 ) + ( 1/3 * 2/3 ) ] * [ ( 1/2 * 1/3 ) + ( 4/5 * 2/3 ) ]] ≈ 0.30

<u />

8 0
3 years ago
Jason wrote 1 1/3 as a sum of 3 fractions. None of the fractions had a denominator of 3. What fractions might Jason have used?
ziro4ka [17]
11/6 , 1/6 and 10/6 are the answer
5 0
2 years ago
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