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4 years ago

Width = (b - a)/n = (8 - 0)/4 = 2

Mathematics

2 answers:

Nookie1986 [14]4 years ago

7 0

Yes, I'm getting C also!

Since it's asking for the left-endpoint Riemann Sum, you will only be using the top left point as the height for each of your four boxes, making -1, -2.5, -1.5, and -0.5 your heights. The bases are all the same length of 2. You don't include f(8) because you're not using right-endpoints, and that would also add another 5th box that isn't included in the 0 to 8 range.

alexgriva [62]4 years ago

3 0

I agree as well. The Riemann Sum diagram will look something like what you see in the attached images. I used GeoGebra to create the graph.

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

40 points i will crown. A, B, C, or D.

VLD [36.1K]

Answer:

C. x > -3

Explanation:

The circle with no filling means it can not be equal

It goes ahead of -3 so it is greater than -3

3 0

3 years ago

A 9.0 ladder rests against the side of a wall. The bottom of the ladder is 1.5 meters from the base of the wall. Determine the m

maria [59]

Well we know that the 9.0 meter ladder is the hypotenuse And it is a right triangle with base of 1.5 meters. we have the adjacent and the hypotenuse. which is cosine. So:
cos (a) = adjacent/hypotenuse
cos (a) = 1.5/9.0
cos (a) = 1 / 6
a = cos^-1 (1 / 6)
a = 80.4 degrees

8 0

3 years ago

Write an equation of a line perpendicular to y = x + 4 in slope-intercept form that passes through the point (-2, 6)

Scrat [10]

$k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\ if\ m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\---------------------\\k:y=x+4;\ l:y=m_2x+b_2\\\\m_1=1\to m_2=-\dfrac{1}{1}=-1\\\\k\ \perp\ l\to l:y=-x+b_2\\\\The\ line\ "l"\ passes\ throught\ the\ point\ (-2;\ 6).\\Subtitute\ x=-2\ and\ y=6\ to\ the\ equation\ of\ the\ line\ "l":\\\\6=-(-2)+b_2\\6=2+b_2\ \ \ |subtract\ 2\ from\ both\ sides\\b_2=4\\\\Answer:\boxed{y=-x+4}$

5 0

4 years ago

The client is to receive cyclophosphamide (Cytoxan) 50 mg/kg intravenously in divided doses over 5 days. The client weighs 176 p

marishachu [46]

Answer:

800 mg

Step-by-step explanation:

As the excerpt states that the client weights 176 pounds and is to receive cyclophosphamide (Cytoxan) 50 mg/kg, we have to convert 176 pounds to kg:

1 kg → 2.20 pounds

x ← 176 pounds

x= (176 pounds * 1 kg)/2.20 pounds= 80 kg

Now, we have to determine the total amount of cyclophosphamide (Cytoxan) that the client has to receive:

50 mg/kg

For 80 kg: 80 kg *50 mg/kg= 4,000 mg

As the dose is divided over 5 days:

4,000 mg / 5= 800 mg

The client will receive 800 mg of cyclophosphamide each day.

6 0

3 years ago