51 is 15% of what number?

Feliz [49]

<h3>Answer: 35</h3>

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Explanation:

Check out the diagram below.

U is between T and V, and U is on segment TV. We can see that the smaller pieces TU and UV add back up to the largest segment TV

TU + UV = TV

17 + 18 = TV

35 = TV

TV = 35

Segment TV is 35 units long.

5 0

2 years ago

Help me out please?<br>chapter : exponents and powers

belka [17]

Answer:

$\frac{125}{27}$

Step-by-step explanation:

$\frac{2^{-4} \times 15^{-3} \times 625}{5^2 \times 10^{-4}}$

Lets expand all the composite numbers into prime numbers.

$=> \frac{2^{-4} \times (3^{-3} \times 5^{-3}) \times 5^4}{5^2 \times (2^{-4} \times 5^{-4})}$

Lets cancel $2^{-4}$ from numerator and denominator.

$=> \frac{3^{-3} \times 5^{-3} \times 5^4}{5^2 \times 5^{-4}}$

Using laws of exponents , lets solve this.

$=> \frac{3^{-3} \times 5^{(-3 + 4)}}{5^{( 2 - 4)}}$

$=> \frac{3^{-3} \times 5^{1}}{5^{-2}}$

$=> 3^{-3} \times 5^{[1 - (-2)]}$

$=> 3^{-3} \times 5^{3}$

$=> \frac{5^3}{3^3} = \frac{125}{27}$

4 0

2 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.

C)

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 <em>instanteous</em> 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 <em>exactly</em> t=5, the rate of growth is 51.72 cells per hour.

And we're done!

7 0

2 years ago

How many days in the month of february??

Rashid [163]

Well it depends if it's a leap year. If it isn't a leap year there are 28 days and if we are in a leap year than there are 29 days in February. Hope that helped :)

7 0

3 years ago

Graph y=5/3x-9 in slope intercept

prisoha [69]

Answer:

Step-by-step explanation:

3 0

2 years ago