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pickupchik [31]

2 years ago

5

A certain type of bacteria reproduces so that after x hours there are 15x cells. What single power represents he size relationsh

ip between the cell population after 24 hours and the cell population after 8 hours? Show your work

1 answer:

algol [13]2 years ago

7 0

Answer:

let's say cells is c and hours are x that means there are 15 cells each hour. the equation for that is c=15x

after 08 hours the cells have multiplied 8 times their original size. meaning that the equation changes to this: c=15x x=8 c= 15(8)

x equals 8 at that time.. literally

ok the cells have multiplied and now there are 120 cells in total

then after one day the cells have multiplied over 12 times afterwards. now x equals 24. the equation would look a little like this:

c=15x x=24 c= 15(24)

there are 360 cells now.

Explain in your own words the relationship between x=8 with x=24 in the equation.

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What is the value of x that makes the given equation true?

Doss [256]

Hello there!

The correct answer is C

-----------------------------

x - 3x = 2(4 + x)
Distribute the right side
x - 3x = (2)(4)+(2)(x)
x - 3x = 8 + 2x
Combine like terms
x - 3x - 2x = 8
-4x = 8
Divide both sides by -4
-4x/4 = 8/4
x = -2

Hence,
The correct answer is c

Have a nice day!

6 0

3 years ago

Read 2 more answers

I need help finding the y-intercept of these, I am no good with slope and tables. ;-;

zubka84 [21]

Answer:

1. -4

2. 6

3. -6

(i hope this is right •_•)

Step-by-step explanation:

<u>First table</u>

When the x coordinate is 0, that point is your y intercept. For the first one, your y intercept is -4.

<u>Second table</u>

For the second one, we need to do a bit of counting. We can see the x coordinates go before and after zero, but the table does not show us 0. So, we're going to have to calculate the slope. Bear with me here, I'm a little tired and this might be rambling ;-;

So, the formula for calculating slope is $\frac{y_{2} -y_{1}}{x_{2}-x_{1}}$ . I'm going to use those first two points to get my slope. We're going to say the first point is point 1 and the second point is point 2 (genius, i know lol). So, let's substitute in the x and y values.

$\frac{9-12}{-3-(-6)}$

And simplify...

$\frac{-3}{3}$

So, your slope is -1. We can use that. We just have to count down.

If the x is -2, the y is 8. If the x is -1, the y is 7. And if the x is 0....(drumroll please) the y is 6! So, your y intercept is 6.

<u>Third table</u>

Now, this table isn't very nice either and doesn't have an x as 0 for us. Hmpf. But we can still do what we did above. I'll just skip the explanation and you can observe my steps.

$\frac{-3-(-5)}{9-3}$

$\frac{2}{6}$

$\frac{1}{3}$

Let's count down....

If x = 3, y = -5

If x = 2, y = -5 1/3

If x = 1, y = -5 2/3

If x = 0, y = -6

8 0

3 years ago

Need steps on how to factor 3x^4 + x^3 + 6x^2 + 2x

ella [17]

Answer:

3x times 3x times 3x times 3x + x^3 + 3x times 2x times 3x times 2x + 2x

or

x^3 + 2x^3 + 3x^6

Step-by-step explanation:

3x^4 + x^3 + 6x^2 + 2x

3x times 3x times 3x times 3x + x^3 + 6x times 6x + 2x

3x times 3x times 3x times 3x + x^3 + 3x times 2x times 3x times 2x + 2x

x^3 + 2x^3 + 3x^6

5 0

2 years ago

Open notes test: Relations, Functions and Slope

Karolina [17]

Answer:

I'm not sure what your asking

6 0

2 years ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

so

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

so

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

so

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago