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elena55 [62]
4 years ago
10

Paulie ordered 250 pens and 250 pencils to sell for a theatre club fundraiser. The pens cost 11 cents more than the pencils. If

Paulie's total order costs $42.50, find the cost of each pen and pencil.
Mathematics
1 answer:
Korolek [52]4 years ago
4 0
Remember that cents has a decimal when put in an equation.
250(x + 11) + 250x = 4250
250x + 2750 + 250x = 4250
500x + 2750 = 4250
500x = 1500
x = 3 cents (pencils)
x+11 = 14 cents (pens)
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marshall27 [118]
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Lacie is filling a bucket with water.
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Answer: Option A) liters

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Liters is the standard unit of measurement for volume. Metres, on the other hand, is the standard unit of measurement for length, height, width or distance traveled.

So, since we are to me measure the amount (volume) of water in Lacie's bucket, liters is the standard unit to used.

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Nate is making a picture frame by cutting a small rectangular prism out of a larger rectangular prism. When the project is compl
dybincka [34]
Surface area= [15*3*2]+[20*3*2]+[10*3*2]+[14*3*2]+[15*3*2]+[14*2.5*2]

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6 0
3 years ago
Which triangle congruence theorem can be used to prove the triangles are congruent
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7 0
3 years ago
Who can help me d e f thanks​
12345 [234]

d)

y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}

Product rule:

y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'

Chain and power rules:

y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'

Power rule:

y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)

Now simplify.

y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

e)

y = \dfrac{3bx + ac}{\sqrt{ax}}

Quotient rule:

y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}

y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}

Power rule:

y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}

Now simplify.

y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}

y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}

y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}

f)

y = \sin^2(ax+b)

Chain rule:

y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'

y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'

y' = 2a \sin(ax+b) \cos(ax+b)

We can further simplify this to

y' = a \sin(2(ax+b))

using the double angle identity for sine.

7 0
2 years ago
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