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ollegr [7]
3 years ago
9

A) Car A travels x km for every litre of petrol while car B travels

Mathematics
1 answer:
ser-zykov [4K]3 years ago
5 0

Answer:   16 litres

Step-by-step explanation:

A:   1l          x  km

B:   1l         x+5   km

Running 400 km   A used   400/x  litres

Running  400 km  B used  400/(x+5)  litres

400/x- 400/(x+5)= 4

400*(x+5)- x*400-4*x*(x+5)=0

400*x+2000-400*x-4*x²-20*x=0

2000-4*x²-20*x=0     difide by 4 both sides of equation

500-x²-5*x=0

Lets solve the equation using discriminant:

D=5²-(-1)*4*500=2025

sqrt(D)=45

x1= (5+45)/(-2)= -25          x2=(5-45)/(-2)=20

x1=-25     x2=20

x1<0 so is not the solution of the problem ( number of litres can't  be negative)

So A uses  litr per x=20 km     and B uses 1 litr per 20+5=25 km

Running 400 km B uses 400/25=16 litres

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Answer:

1. No

2. Yes

3. Yes

4. Yes

Step-by-step explanation:

In 1. the difference between each term is not constant.(it goes up by, 2, 4, 6)

2. The terms increase by the same amount(it all goes up by 3)

3. The terms decrease by the same amount(it all goes down by 5)

4. The terms increase by the same amount(it all goes up by 6)

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3 years ago
Find the distance between the points 10,2 and 5,-10
melomori [17]
The distance between the points is 13
3 0
2 years ago
In the equation 9^2 x 27^3 = 3^x, what is the value of x? <br> Show your work on scratchpaper
IceJOKER [234]

Answer:

x=13

Step-by-step explanation:

9^2 * 27^3 = 3^x

We need to get each term with a base of 3

9^2 = (3^2) ^2

We know that a^b^c = a^(b*c)

(3^2) ^2 = 3^(2+2) = 3^4

27^3 = (3^3) ^3 = 3^(3*3) = 3^9

Replacing these in the original equation

3^4 * 3^9 = 3^x

We know that a^b *a^c = a^(b+c)

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The bases are the same, so the exponents must be the same

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7 0
3 years ago
In a jewelry store, rings makeup 5/9 of the inventory. Earrings make up 4/15 of the inventory. How many times greater is the rin
Free_Kalibri [48]
First let's make the denominators equal, so that we can compare the two fractions more easily.
To do this, we have to find a number that both 15 and 9 divide into. The smallest number that this can happen to is 45.

15 x 3 = 45
9 x 5 = 45

So we have to multiply 15 by 3 in order to make it into 45. If we're changing the denominator, the numerator must change too, by the same multiplier.

4 ---> 4 x 3 = 12
--
15 ---> 15 x 3 = 45

4/15 = 12/45


5 ---> 5 x 5 = 25
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9 ---> 9 x 5 = 45

5/9 = 25/45

To find how many times more rings there are, we divide 25 by 12. As the result of this isn't an integer, we leave the answer as a fraction:

There rings inventory is 25/12 times bigger than the earrings inventory.
4 0
3 years ago
Can some answer pls.
kirill [66]

Answer:

E. (7,-2)

Step-by-step explanation:

5 0
3 years ago
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