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Dovator [93]
3 years ago
11

The number of cubic yards of dirt, D

Mathematics
1 answer:
Sholpan [36]3 years ago
8 0
What is the question exactly?
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At what time since leaving their flowers do Bee 1 and Bee 2 have the same distance remaining?
GrogVix [38]

By reading the given graph with the two linear functions, we want to see at which time do the two bees have the same distance remaining. We will see that the correct option is B, 6 minutes.

So, in the graph, we have distance remaining on the vertical axis and time on the horizontal axis.

We also have two lines, each one describing the distance of each bee as a function of time.

We want to see at which time do the two bees have the same distance remaining, thus, we need to see when the lines intersect (this means that for the same time, the two bees have the same distance remaining).

In the graph, we can see that the intersection happens at the time of 6 minutes, thus the correct option is B; 6 minutes.

If you want to learn more about linear function's graphs, you can read:

brainly.com/question/4025726

6 0
2 years ago
Rachel's learning to play the violin Monday through Friday she practices 45 minutes on each day on Saturday and Sundays he broug
I am Lyosha [343]

Rachel spends 6 hours 45 minutes a week trying to learn to play the violin.

Step-by-step explanation:

Step 1; Rachel learns for 45 minutes a day from Monday through Friday and in the weekend she learns to play the violin for one and a half hours. So she learns for the following periods of time

Monday - 45 minutes

Tuesday - 45 minutes

Wednesday - 45 minutes

Thursday - 45 minutes

Friday - 45 minutes

Saturday - 90 minutes (60 × 1.5 hours)

Sunday - 90 minutes (60 × 1.5 hours)

Step 2; To determine how much time she practices in a week we just add the individual times she plays on each day.

Total time practices in a week = 45 + 45 + 45 + 45 + 45 + 90 + 90 = 405 minutes = 6 hours 45 minutes.

           

5 0
2 years ago
Can someone help please ? Thanks I’m advance
Pani-rosa [81]

Answer:

f(-3) = 16

Step-by-step explanation:

Given:

f(x) = 2x² + x + 1

Required:

f(-3)

Solution:

Substitute x = -3 into f(x) = 2x² + x + 1

f(-3) = 2(-3)² + (-3) + 1

f(-3) = 2*9 - 3 + 1

f(-3) = 18 - 3 + 1

f(-3) = 16

6 0
3 years ago
When x = 2, y = 50 and when x = 4, y = 100. Which direct variation equation can be used to model this function?
Karolina [17]
Well, let's first write these as points.

( 2 , 50 )

( 4 , 100 )

We can see that when "x" is reduced by 2, "y" is reduced by 50. This means that if we reduce "x" by 1, "y" will be reduced by 25. Thus, we can say that if "x" is 1, "y" will be 25.

( 1 , 25 )

What we know that 25 * 1 = 25, and that 2 * 25 = 50. We can see that multiplying "x" by 25 will give us our "y". We can now write this as an equation.

y = 25x
3 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
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