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Help me pls 20 POINTS

Tema [17]

Answer:

bottom =80m^2

Side =50m^2

triangle = 3 ×8 ×5

6 0

2 years ago

What is the value of x in the equation 4(2x 10) = 0? −5 −7 5 7

Alex73 [517]

4(2x+10)= 0
8x+40=0
subtract 40 from both sides
8x+40-40=0-40
8x=-40
x=-40/8
x=-5

4(2x-10)=0
8x-40=0
Add 40 from both sides
8x-40+40=0+40
8x=40
x=40/8
x=5

4 0

2 years ago

Read 2 more answers

An item costs $64. The price is increased by 10%, then

ddd [48]

Yes because you are basically taking what you are giving in the first place

6 0

2 years ago

A biologist is studying the growth of a particular species of algae. She writes the fall inclusion to show the radius of the alg

horrorfan [7]

Answer:

Part A)

A reasonable domain to plot the growth function is:

$0\leq d \leq 12$

Part B)

The y-intercept represents that when the biologist started her study, the radius of the algae was five millimeters.

Part C)

The average rate of change from d = 4 to d = 11 was about 0.11. This means that from the 4th day to the 11th, the radius of the algae grew, on average, at a rate of 0.11 mm per day.

Step-by-step explanation:

The radius of the algae f(d) in millimeters after d days is given by the function:

$f(d)=5(1.02)^d$

Part A)

We know that the radius of the algae was approximately 6.34 mm when the biologist concluded her study. To find the reasonable domain, we can substitute 6.34 for f(d) and solve for d. Therefore:

$6.34=5(1.02)^d$

Divide both sides by five:

$(1.02)^d=1.268$

Take the log of both sides with base 1.02:

$\displaystyle d=\log_{1.02}1.268$

Using the Change of Base Property, evaluate for d:

$\displaystyle d=\frac{\log 1.268}{\log 1.02}=11.9903...\approx 12$

So, the biologist concluded her study after 12 days.

Therefore, a reasonable domain to plot the growth function is:

$0\leq d\leq 12$

Part 2)

The y-intercept of the function is when d = 0. Find the y-intercept:

$f(0)=5(1.02)^{(0)}=5(1)=5$

Since d represent the amount of days after the study had begun, the y-intercept represents the radius of the algae on the initial day.

So, when the biologist started her study, the radius of the algae was five millimeters.

Part 3)

To find the average rate of change for a nonlinear function, we find the slope between the two endpoints on the interval.

We want to find the average rate of change of f(d) from d = 4 to d = 11.

Find the endpoints:

$f(4)=5.4121...\text{ and } f(11)=6.2168...$

And find the slope between them:

$\displaystyle m=\frac{f(11)-f(4)}{11-4}=0.1149...\approx 0.11$

Since f(d) measures millimeters and d measures days, this tells us that, on average, the radius of the algae grew by about 0.11 mm per day from the 4th day to the 11th day.

5 0

2 years ago

Determine the maximum value of the objective function, P.

Ad libitum [116K]

Answer:

Maximum (250,125) Answer

Step-by-step explanation:

This is more of a graphing problem than it is anything else.

Begin by graphing all 4 given equations.

When you do that, mark the intersection points of at least 2 lines. In this case it is exactly 2 lines for each intersecting point.

6x + 4y <= 2000 and 2x + 4y <= 1000 intersect at (250,125)

x=>0 and y=>0 intersect at (0,0)

6x + 4y <=2000 and x => 0 intersect at 333.333

2x + 4y <=1000 and y>=0 intersect at 0,250.

Any other intersection points fall outside the range of the givens. The shaded part we are interested in is sort of a very dark green/blue. It is the interior of the quadrilateral determined by the 4 vertices that are marked.

Now all you have to do is determine the maximum point using P=15x + 12y

For 0,0 P = 15*0 + 12,0 = 0

For 0,250 P = 15*0 + 12*250 = 3000

For 333.3333,0 P = 15*333.3333 + 12*0 = 5000 rounded.

For 250,125 P = 15*250 + 12*125 = 5250 Which is the maximum

Answer (250,125) produces the maximum value Answer

5 0

3 years ago