In a college biology class, only half of the class earned a passing grade, Out of all the students

Use the following graph to estimate the rate of change of the function at x=0.6 using the points (0,0) and (1,−0.5)

svetoff [14.1K]

Answer:

The rate of change of the function at x = 0.6 is approximately -0.72

Step-by-step explanation:

The given information from the graph are;

Points (0, 0) and (1, -0.5)

We have that the graph is that of a cubic function, therefore;

f(x) = ax³ + bx² + cx + d

At f'(x) at (0, 0) and (1, -0.5) = 0 gives;

3ax² + 2bx + c = 0

∴ c = 0

Also

ax³ + bx² + cx + d = 0

d = 0

We have

3a(1)² + 2b(1) + 0 = 0

3·a + 2·b = 0..............(1)

a(1)³ + b(1)² + c(1) + d = -0.5

a + b + 0×(1) + 0 = -0.5

a + b = -0.5..........(2)

Multiplying equation (2) by 2 and subtracting it from equation (1) gives;

2 × (a + b = -0.5) = 2·a + 2·b = -1

3·a + 2·b - (2·a + 2·b) = 0 - (-1) = 0 + 1 = 1

a = 1

From

a + b = -0.5, we have;

1 + b = -0.5 = 0

b = -0.5 - 1 = -1.5

The equation becomes

f(x) = x³ - 1.5·x²

The rate of change of the function at x = 0.6 is therefore given as follows;

f'(x) = 3 × x² - 2 × (1.5)·x = 3·x² - 3·x

At x = 0.6, we have;

f'(0.6) = 3·(0.6)² - 3·(0.6)

f'(0.6) = 3×0.6² - 3×0.6 = -0.72

The rate of change of the function at x = 0.6 ≈ -0.72

4 0

3 years ago

For y = 6/x² estimate the value of 6/(4.03)².

Aleks [24]

=ligma baalls so that I can tape this d to your fore head so you can cd’s nuts

6 0

2 years ago

The population of a certain community is known to increase at a rate proportional to the number of people present at time t. If

attashe74 [19]

Answer:

$t=7.85 years$

Step-by-step explanation:

We can write this rate as an ordinary differential equation.

$\frac{dP}{dt}=aP$

Where a is proportional constant, P the population variable, and t the time.

$\frac{dP}{P}=adt$

Integrating each side of the equation.

$\int \frac{dP}{P}=\int adt$

$ln(P)=at+c$

$P=e^{at+c}$

$P=e^{c}e^{at}=Ce^{at}$

To find C we need to use the initial condiction, it means evaluae P at t=0.

$P_{0}=Ce^{0}=C$

$P=P_{0}e^{at}$

Now, we use the sentence the population has doubled in 5 years.

$2P_{0}=P_{0}e^{a5}$

We can find "a" in this condition.

$2=e^{a5}$

$ln(2)=a5$

$a=\frac{ln(2)}{5}$

$a=0.14$

Finally, let's find how long will it take to triple.

$3P_{0}=P_{0}e^{0.14t}$

$3=e^{0.14t}$

$t=\frac{ln(3)}{0.14}$

$t=7.85 years$

I hope it helps you!

4 0

2 years ago

Andrea, Justin, and Virginia went to an office supply store. Andrea bought 6 pencils, 7 markers, and 8 erasers. Her total was $1

olga2289 [7]

Answer:

$x=0.25\,,\,y=1.75\,,\,z=0.25$

Step-by-step explanation:

Let $x,y,z$ denotes number of pencils, markers, erasers respectively.

For Andrea:

$6x+7y+8z=15.75\,\,\,(i)$

For Justin:

$6x+8y+5z=16.75\,\,\,(ii)$

For Virginia:

$5x+5y+7z=11.75\,\,\,(iii)$

On subtracting equations (i) and (ii), we get

$(6x+7y+8z)-(6x+8y+5z)=15.75-16.75\\-y+3z=-1\\y-3z=1\\y=1+3z$

Put $y=1+3z$ in equation (i)

$6x+7(1+3z)+8z=15.75\\6x+29z=8.75\,\,\,(iv)$

Put $y=1+3z$ in equation (iii)

$5x+5(1+3z)+7z=11.75\\5x+22z=6.75\,\,\,(v)$

Multiply equation (iv) by 5 and equation (v) by 6 and then subtract both the equations.

$5(6x+29z)-6(5x+22z)=5(8.75)-6(6.75)\\13z=3.25\\z=\frac{3.25}{13}=0.25$

Put $z=0.25$ in equation $y=1+3z$

$y=1+3(0.25)=1.75$

Put $y=1.75\,,\,z=0.25$ in equation (i)

$6x+7(1.75)+8(0.25)=15.75\\6x+12.25+2=15.75\\6x=1.5\\x=0.25$

8 0

3 years ago

Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

Art [367]

Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$

$\text{The length of the longer diagonal}=4+9.54$

$\text{The length of the longer diagonal}=13.54$

Therefore, the length of longer diagonal is approximately 13.54 inches.

3 0

3 years ago