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2 years ago

14

I need to find the slope, x intercept and y intercept please help

1 answer:

Sati [7]2 years ago

6 0

Solve for y. the slope will be m, y intercept will be b

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Solve for x, thank you

Arisa [49]

Answer:

10.80

Step-by-step explanation:

tan42=x/12

x=12tan42=10.80

3 0

2 years ago

What is 9/12 in simplest form

olga nikolaevna [1]

Solutions

⇒9/12

1) To see if 9/12 is in simplest form you have to find 9 and 12's GCF.
The GCF of 9 and 12 is 3

2) Divide by the GCF
9 ÷ 3 = 3
12 ÷ 3= 4

3/4 is answer

6 0

3 years ago

Read 2 more answers

3, 10, 17, 24. what is the 50th term in the sequence

Kamila [148]

Check if you can write an equation relating the term number to the actual value
n1=3
n2=10 = 3+7
n3= 17 = n2+7 = n1+7+7 = n1 +2*7
n4= 24 = n1+3*7

so you will notice a pattern
for the x-th term

n_x =3+(x-1)*7

the 50th term would be n_50 = 3+(50-1) * 7

4 0

2 years ago

PLZ HELP!!!! WILL GIVE BRAINLIEST + WORTH 10 POINTS!!! THX!!!!

mafiozo [28]

Answer:

I got this answer hope works...

5 0

3 years ago

Read 2 more answers

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