2 years ago

−5(1−5k)−4(2k+5), I got this problem on khan academy and can't figure it out

Mathematics

1 answer:

ikadub [295]2 years ago

3 0

Your answer would be 1.13 repeating or no solution.

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How many ways can a committee chairman choose 4 people for a subcommittee if there are 12 members on the whole committee?

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A group of k elements can be chosen from a group of n elements in $\frac{n!}{k!(n-k)!}$ ways.

$(^{12} _4)=\frac{12!}{4!(12-4)!}=\frac{12!}{4! \times 8!}=\frac{8! \times 9 \times 10 \times 11 \times 12}{2 \times 3 \times 4 \times 8!}=\frac{9 \times 10 \times 11 \times 12}{2 \times 3 \times 4}= \\ \\
=3 \times 5 \times 11 \times 3=495$

The answer is B. 495.

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Alex17521 [72]

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2 years ago

3x2 − y3 − y3 − z if x = 3, y = −2, and z = −5.

tatyana61 [14]

Ok so from experience on this site
3x2=3x^2
people don't know how to put exponents
3x^2-y^3-y^3-z
simplify first

3x^2-2y^3-z

x=3
y=-2
z=-5
3(3^2)-2(-2)^3-(-5)
3(9)-2(-8)+5
27+16+5
48

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3 years ago