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AURORKA [14]
3 years ago
8

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)
coldgirl [10]

Answer:

X equals to -11

Step-by-step explanation:


6 0
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
8 0
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.

4 0
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
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A= l x w
5x * 2x^2 - 4x + 13
10x^3 - 20x^2 + 65x
3 0
3 years ago
Read 2 more answers
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