3 years ago

Please help me on number 11

Mathematics

1 answer:

AysviL [449]3 years ago

4 0

Answer:

41.1

Step-by-step explanation:

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Answer:

there was 3 containers left with cup each of hot fudge

Step-by-step explanation:

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4 years ago

PLS SHOW UR SOLUTIONS

maksim [4K]

Answer:

Q1. $(2x-3)^{2}+3(x-4)=5+y\\$

$4x^{2} -12x+9+3x-12-5-y=0$

$4x^{2} -9x-8-y=0$

$4x^{2} -9x-8=y$

∴ $y = 4x^{2} -9x-8$

∴ From the quadratic equation above, a = 4, b = -9, and c = -8.

Q2. $(x-5)^{2} -x^{2} =y$

$x^{2} -10x+25-x^{2} =y$

$-10x+25=y$

∴ $y = -10x + 25$

∴ The equation above is not a quadratic equation as it does not follow the format of ax² + bx + c.

Q3. $2x(4x^{2} -4x+1)=y+x(8x)+4(2x)$

$8x^{3} -8x^{2} +2x-8x^{2} -8x=y$

∴ $y=8x^{3} -16x^{2} -6x$

∴ The equation above is not a quadratic equation as it does not follow the format of y = ax² + bx + c.

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1 year ago

Which of the following sets of numbers could not represent the three sides of a right

blondinia [14]

Answer:

The answer is (40, 42, 59)

Step-by-step explanation:

This is because when you use pythagoras theorem by adding the squares of 40 & 42 and then square root the answer you wouldn't get 59

Rather you will get 58

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3 years ago

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Nataly [62]

Answer:

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3 years ago

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3 years ago