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3 years ago

Does the table show a proportional relationship? If so, what is the value of y when x is 11

Mathematics

2 answers:

prisoha [69]3 years ago

8 0

Answer:

Yes, the table shows a proportional relationship

The value of y when x is 11 is 1331

Step-by-step explanation:

Let us check the relation between x and y in the table

∵ x = 4 and y = 64

∵ 64 = 4³

∴ y = x³

∵ x = 5 and y = 125

∵ 125 = 5³

∴ y = x³

∵ x = 6 and y = 216

∵ 216 = 6³

∴ y = x³

∵ x = 10 and y = 1000

∵ 1000 = 10³

∴ y = x³

∵ All the values on the table give the same relation

∴ x and y are proportion

∴ The table shows a proportional relationship

∵ y = x³

∵ x = 11

∴ y = (11)³

∴ y = 1331

∴ The value of y when x is 11 is 1331

storchak [24]3 years ago

8 0

Answer:

Step-by-step explanation yeah

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Two teams are distributing information booklets. Team A distributes 60% more boxes of booklets than Team B, but each box of Team

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Answer:

C. 4,100

Step-by-step explanation:

"60% more" is represented by a multiplier of 1 + 0.60 = 1.60.

"60% fewer" is represented by a multiplier of 1 - 0.60 = 0.40.

Let b represent the number of booklets distributed by Team B. Then the number distributed by Team A is ...

1.60 × (0.40b) . . . . 60% more boxes, each with 60% fewer booklets

= 0.64b

Then the total distributed by both teams is ...

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The only answer choice that is a multiple of 41 is ...

4,100 . . . choice C

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3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$