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Margaret [11]
3 years ago
9

(a) has no solutions.(b) has infinitely many solutions.(c) has exactly one solution.

Mathematics
1 answer:
grandymaker [24]3 years ago
6 0

Answer and Step-by-step explanation:

Given

ax + by = c

qx + ry = s

(a) the equation has no solutions if a/q = b/r ≠ c/s, when this happens, we say the system of equations has no solution. For example

x + y = 3

x + y = 5

Subtracting first equation from the second we have:

0 = 2 which is impossible.

(b) the equations have infinite solutions if a/q = b/r = c/s, for example

x + y = 2

x + y = 2

Subtracting the first equation from the second we have

0 = 0, since this is always true, then it has infinite solutions.

(c) the equations have unique solutions if a/q ≠ b/r, for example

x + y = 2

x – y = 1

Adding the first and second equation we have

2x = 3, we can get x from here and definitely y, so we have just one solution.

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lakkis [162]
The weight of the hamburger is 160g after two bites.

Each bite takes away 20% of the weight of the burger, so we'll divide the initial weight by a percentage to find the weight before the bite.

The starting weight is 160g. Because 20% of the weight was removed from a bite, we'll divide this weight by the percentage of the burger remaining after the bite:

100 - 20 = 80

80% of the burger is left after a bite. Convert this fraction into a decimal by dividing by 100:

80 \div 100 = 0.8

Divide the initial weight by this decimal:

160 \div 0.8 = 200

The weight of the burger before the second bite was 200.

We'll do the same procedure for the first bite. The initial weight is 200, and there is 80% of the burger remaining after the first bite. Because 80% is 0.8 in decimal form, divide 200 by 0.8:

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The starting weight of the burger before the bites was 250 grams.
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Hope this helps!

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