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3 years ago

Y = 9/7x + 8/5 Find the y-intercept and simplify it.

Mathematics

1 answer:

Doss [256]3 years ago

6 0

Answer:

1.6

Step-by-step explanation:

y-intercept, x=0

y=0+8/5

y=8/5

=1.6

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Solve for x<br> 5(x + 8) = 2x – 7 + 8(x - 1)

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Answer:

X equals to -11

Step-by-step explanation:

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3 years ago

How do you answer this equation?<br> 5a - 12 = 48

IgorC [24]

Add 12 to both sides resulting in 5a = 60 and then divide both sides by 5 which results in a = 12 so your answer will be a = 12

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2 years ago

The table and the graph each show a different relationship between the same two variables, x and y:

Gelneren [198K]

Answer:

110

Step-by-step explanation:

The answer is 110

If the function for the table is y=90x and the function for the graph is y=80x, then at x=11, the table would be at (11, 990) and the graph at (11, 880) meaning that the y value of the table would be 110 more than the y value of the graph.

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2 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

2 years ago

The length of a rectangle is represented by the function L(x) = 5x. The width of that same rectangle is represented by the funct

IgorC [24]

A= l x w
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10x^3 - 20x^2 + 65x

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3 years ago

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