2 years ago

Please help right now??

Mathematics

2 answers:

makvit [3.9K]2 years ago

8 0

The picture is blank bsnsbsbsbdbbdbshd

timurjin [86]2 years ago

5 0

The picture is black HAHAH

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Given the hyperbola y = 1/x, find the area under the curve between x = 5 and x = 33 to 3 sig. dig. namely: integral subscript 5

Shtirlitz [24]

Answer:

(b) ln(33/5)

Step-by-step explanation:

$\int\limits_{5}^{33} \frac{1}{x} \, dx = \ln(x) \Big|\limits_{5}^{33} = \ln(33) - \ln(5) \\\\= ln(33/5)$

Remember that in general

$\ln(a) - \ln(b) = \ln(a/b)$

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2 years ago

I need help I can’t figure it out

dusya [7]

Answer:

$\large\boxed{\dfrac{41p}{70q^4(r-9)^2}}$

Step-by-step explanation:

$\dfrac{2p^4}{5q^5(r-9)^3}\cdot\dfrac{41q(r-9)}{28p^3}\\\\=\dfrac{2\!\!\!\!\diagup^1\cdot41}{5\cdot28\!\!\!\!\!\diagup_{14}}\cdot\dfrac{q\!\!\!\!\diagup}{q^{5\!\!\!\!\diagup^4}}\cdot\dfrac{(r\!\!\!\!\!\!{--}-9)\!\!\!\!\!\!{--}}{(r-9)^{3\!\!\!\!\diagup^2}}\cdot\dfrac{p^{4\!\!\!\!\diagup}}{p^3\!\!\!\!\!\!\diagup}\\\\=\dfrac{41}{70}\cdot\dfrac{1}{q^4}\cdot\dfrac{1}{(r-9)^2}\cdot\dfrac{p}{1}\\\\=\dfrac{41p}{70q^4(r-9)^2}$

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2 years ago

Solve 2(4 x + 3) < 5 x + 21. { x | x < 9} { x | x > -5} { x | x > -9} { x | x < 5}

Nesterboy [21]

Answer:

x <5

Step-by-step explanation:

2(4 x + 3) < 5 x + 21

Distribute

8x +6 < 5x +21

Subtract 5x from each side

8x -5x +6 < 5x-5x+21

3x +6 < 21

Subtract 6 from each side

3x+6-6 <21-6

3x <15

Divide each side by 3

3x/3 <15/3

x <5

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2 years ago

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kupik [55]

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Step-by-step explanation:

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3 years ago

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11. On Halloween, 87.5% of the homes in a

fiasKO [112]

87.5% of 104
87.5% x 104 = 91 homes handed out candy

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2 years ago

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