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3 years ago

13y-7y=9 Solve for y, write it as a reduced fraction.

Mathematics

1 answer:

raketka [301]3 years ago

4 0

Answer:

3/2 OR 1 1/2

Step-by-step explanation:

Collect like terms, meaning combine both y's. You should have something like this afterwards.

6y = 9

Now that you have that divide the 6 on both sides so Y is by itself!

Your answer is now 3/2.

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The answer is SSS

Step-by-step explanation:

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4 years ago

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Solve 2(x + 3) = x - 4

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Answer:

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Step-by-step explanation:

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3 years ago

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Figure A is a scale image of figure B Figure A maps to figure B with a scale factor of 4/9 What is the value of x?

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16 divided by 4/9= 36 which is your answer

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Tyler knows that the distance around a running track is 400 meters. Beth, Tyler's sister, decides to measure the running track w

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Answer:

Step-by-step explanation:

True measurements = 400 meters

Observed measurements = 392 meters

% error = change in measurements / true measurements × 100/1

= (400 - 392) / 400 × 100

= 8/400 × 100

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3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago