Ask question

Ask question

All categories

antoniya [11.8K]

3 years ago

9

I dont understand plz help

2 answers:

mylen [45]3 years ago

5 0

Use pemdas

Parenthesis first
Then exponents
Multiplication
Division
Addition
And then subtraction

ddd [48]3 years ago

4 0

Here is answer in the picture I hope this is help to you

You might be interested in

What is the sum of 1 2/16 + .6 + (-½)

Mashutka [201]

Answer:

1 9/40

Step-by-step explanation:

Any calculator can tell you the sum is ...

1 + 2/16 + 0.6 - 1/2 = 1.225

The decimal fraction can be reduced, if you like:

225/1000 = (25)(9)/(25(40)) = 9/40

As a mixed number, the sum is 1 9/40.

__

Alternate solution

1 2/16 + 0.6 -1/2 = 1 1/8 + (6/10 -5/10) = 1 1/8 + 1/10

= 1 + (1/8 +1/10) = 1 +(5/40 +4/40) = 1 9/40

6 0

2 years ago

An art class is making a mural for their school which has a triangle drawn in the middle. The length of the bottom of the triang

Feliz [49]

Answer:

perimeter = 6x+15

Step-by-step explanation:

7 0

2 years ago

The initial quantity at T= 0 is 1000z the quantity grows by a factor 3%. What is the quantity at T=5?

fgiga [73]

Answer:

1,159

Step-by-step explanation:

1000*(1.03) = 1159.274 at t = 5

8 0

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

Find the Value of 2a2+5b2

Valentin [98]

Answer:

I get 92 but not really sire

3 0

3 years ago