All categories

2 years ago

Can you please help me...Thank you!

Mathematics

1 answer:

ZanzabumX [31]2 years ago

8 0

Answer:

the answer is the last one.

You might be interested in

A 78 gram sample of Uranium loses half of its mass each year. What is the exponential equation? Please explain why A is the corr

Fiesta28 [93]

Answer:

Step-by-step explanation:

The first answer is correct because we have a decay factor.

The sample is losing mass, so the number that is being multiplied by a power of x must be less than 1.

If the second answer were used, then the sample would be gaining mass.

8 0

2 years ago

Www-awn.aleks.com/alekscgi/x/Isl.exe/1o_u-IgNsikas

mel-nik [20]

Answer:

Step-by-step explanation:

404 not found

4 0

2 years ago

X = ? Find the composite function 20 points and brainliest

masha68 [24]

Answer:

Explanation:

fg(x) = (x+2)(3x^2 - 1)
= 3x^3 - x + 6x^2 - 2
= 3x^3 + 6x^2 - x - 2

gf(x) = (3x^2 - 1)(x + 2)
= 3x^3 + 6x^2 - x - 2

As you can see:

fg(x) = gf(x)
3x^3 + 6x^2 - x - 2 = 3x^3 + 6x^2 - x - 2

5 0

2 years ago

How many terms are in the simplified expression -3x 2 + 2y 2 + 5xy - 2y + 5x 2 - 3y 2

Vilka [71]

there are 6 terms in the simplified expression

4 0

3 years ago

Read 2 more answers

50 points! I understand A. and B. but I would really appreciate help with C.

FromTheMoon [43]

Answer:

$51.72\text{ cells per hour}$

Step-by-step explanation:

So, the function, P(t), represents the number of cells after t hours.

This means that the derivative, P'(t), represents the instantaneous rate of change (in cells per hour) at a certain point t.

So, we are given that the quadratic curve of the trend is the function:

$P(t)=6.10t^2-9.28t+16.43$

To find the instanteous rate of growth at t=5 hours, we must first differentiate the function. So, differentiate with respect to t:

$\frac{d}{dt}[P(t)]=\frac{d}{dt}[6.10t^2-9.28t+16.43]$

Expand:

$P'(t)=\frac{d}{dt}[6.10t^2]+\frac{d}{dt}[-9.28t]+\frac{d}{dt}[16.43]$

Move the constant to the front using the constant multiple rule. The derivative of a constant is 0. So:

$P'(t)=6.10\frac{d}{dt}[t^2]-9.28\frac{d}{dt}[t]$

Differentiate. Use the power rule:

$P'(t)=6.10(2t)-9.28(1)$

Simplify:

$P'(t)=12.20t-9.28$

So, to find the instantaneous rate of growth at t=5, substitute 5 into our differentiated function:

$P'(5)=12.20(5)-9.28$

Multiply:

$P'(5)=61-9.28$

Subtract:

$P'(5)=51.72$

This tells us that at exactly t=5, the rate of growth is 51.72 cells per hour.

And we're done!

7 0

2 years ago