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TiliK225 [7]
1 year ago
11

A population of insects increases at a rate of 1. 5% per day. About how long will it take the population to double? 2. 5 days 5.

0 days 46. 6 days 66. 7 days.
Mathematics
1 answer:
torisob [31]1 year ago
4 0
To calculate the Population Growth (PG) we find the difference (subtract) between the initial population and the population at Time 1, then divide by the initial population and multiply by 100.
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Kyle is having a party he has 43 hot dogs there will be 8 people at his party how many hotdogs will each person get
Brrunno [24]

each person will get 5 hot dogs.

Step-by-step explanation:

Total number of Hot dogs =43

Total number of people at party= 8

Each person will get hot dogs= \frac{Total number of Hot dogs}{Total number of people at party }

Each person will get hot dogs= \frac{43}{8}

Each person will get hot dogs= 5.37 or ≈ 5

So, each person will get 5 hot dogs.

Keywords: Word Problems

Learn more about Word Problems at:

  • brainly.com/question/10978510
  • brainly.com/question/11007026
  • brainly.com/question/10717746

#learnwithBrainly

6 0
3 years ago
Find the exact value of cos theta​, given that sin thetaequalsStartFraction 15 Over 17 EndFraction and theta is in quadrant II.
vova2212 [387]

Answer:

cos \theta = -\frac{8}{17}

Step-by-step explanation:

For this case we know that:

sin \theta = \frac{15}{17}

And we want to find the value for cos \theta, so then we can use the following basic identity:

cos^2 \theta + sin^2 \theta =1

And if we solve for cos \theta we got:

cos^2 \theta = 1- sin^2 \theta

cos \theta =\pm \sqrt{1-sin^2 \theta}

And if we replace the value given we got:

cos \theta =\pm \sqrt{1- (\frac{15}{17})^2}=\sqrt{\frac{64}{289}}=\frac{\sqrt{64}}{\sqrt{289}}=\frac{8}{17}

For our case we know that the angle is on the II quadrant, and on this quadrant we know that the sine is positive but the cosine is negative so then the correct answer for this case would be:

cos \theta = -\frac{8}{17}

5 0
2 years ago
There are two consecutive positive even integers such that the square of the first is 364 more than five times the second. What
xxMikexx [17]

Answer:

Step-by-step explanation:

There are two consecutive positive even integers such that the square of the first is 364 more than five times a second. What are the two numbers?

Two consecutive year positive integers are represented by

x and x + 1

First integer = x

Second integer = y

There are two consecutive positive even integers such that the square of the first is 364 more than five times a second.

This is represented mathematically as:

x² = 364 + 5(x + 1)

x² = 364 + 5x + 5

x² -5x -5 - 364

x² - 5x - 369

6 0
2 years ago
An equation for <br> (-6,-3) (6,5)
Delvig [45]
Y= 2/3x+1
or
y=2/3x+b
.....................
6 0
3 years ago
If x/y + y/x = -1 , find the value of x^3 - y^3
marin [14]

Answer:

  0

Step-by-step explanation:

Multiplying the first equation by xy, we have ...

  x^2 +y^2 = -xy

Factoring the expression of interest, we have ...

  x^3 -y^3 = (x -y)(x^2 +xy +y^2)

Substituting for xy using the first expression we found, this is ...

  x^3 -y^3 = (x -y)(x^2 -(x^2 +y^2) +y^2) = (x -y)(0) = 0

The value of x^3 -y^3 is 0.

7 0
3 years ago
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