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vlabodo [156]
3 years ago
9

Please please help please please help me please please help please please

Mathematics
1 answer:
USPshnik [31]3 years ago
5 0

Answer:

Q2) A 0.5//// B 0.11//// C 1.7//// D 1.15//// E 2.13//// F 4.1

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A ball is thrown from an initial height of 3 meters with an initial upward velocity of 13 m/s. The ball's height h
zlopas [31]

Answer:

the end

Step-by-step explanation:

3 0
3 years ago
Which of the following numbers has a value that is between 10% and 1/9? (a) 0.151 (b) 0.112 (c) 0.108 (d) 0.019
Westkost [7]
First you have to change each one into a decimal.

10% = 0.10
1/9 = 0.11

So the answer has to be between 0.10 and 0.11.

A is more than 0.11.
B is more than 0.11.
C is between 0.10 and 0.11.
D is less than 0.10.

The answer is C. 0.108.
6 0
3 years ago
Given that NML is similar to SQR, what is x?<br><br> a) x=18<br> b) x=14.2<br> c) x=40.5<br> d) x=24
Anni [7]

a) x=18

16÷24 = 2/3

2/3 × 27=18

8 0
3 years ago
How do you do the last two questions?
Mashutka [201]

Answer:

P = 6200 / (1 + 5.2e^(0.0013t))

increases the fastest

Step-by-step explanation:

dP/dt = 0.0013 P (1 − P/6200)

Separate the variables.

dP / [P (1 − P/6200)] = 0.0013 dt

Multiply the left side by 6200 / 6200.

6200 dP / [P (6200 − P)] = 0.0013 dt

Factor P from the denominator.

6200 dP / [P² (6200/P − 1)] = 0.0013 dt

(6200/P²) dP / (6200/P − 1) = 0.0013 dt

Integrate.

ln│6200/P − 1│= 0.0013t + C

Solve for P.

6200/P − 1 = Ce^(0.0013t)

6200/P = 1 + Ce^(0.0013t)

P = 6200 / (1 + Ce^(0.0013t))

At t = 0, P = 1000.

1000 = 6200 / (1 + C)

1 + C = 6.2

C = 5.2

P = 6200 / (1 + 5.2e^(0.0013t))

You need to change the exponent from negative to positive.

The inflection points are where the population increases the fastest.

5 0
3 years ago
What’s the domain and range of:<br> log(√(2x-1) + 3 )<br> Please explain how you got it too!!
Radda [10]

Two main facts are needed here:

1. The logarithm \log x, regardless of the base of the logarithm, exists for x>0.

2. The square root \sqrt x exists for x\ge0.

(in both cases we're assuming real-valued functions only)

By (2) we know that \sqrt{2x-1} exists if 2x-1\ge0, or x\ge\dfrac12.

By (1), we know that \log(\sqrt{2x-1}+3) exists if \sqrt{2x-1}+3>0, or \sqrt{2x-1}>-3. But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require x\ge\dfrac12, so the function has domain \left[\dfrac12,\infty).

To determine the range, we need to know that, in their respective domains, \sqrt x and \log x increase monotonically without bound. We also know that x=\dfrac12 at minimum, at which point the square root term vanishes, so the least value the function takes on is \log3. Then its range would be [\log3,\infty).

3 0
3 years ago
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