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KengaRu [80]
2 years ago
14

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

Mathematics
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
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