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EleoNora [17]
3 years ago
11

The tables show functions representing the growth of two types of bacteria on certain days within an experiment that lasted a to

tal of 10 days. How do the functions in the table compare? A. Since x-intercepts indicate the amount of each bacteria at the start of the experiment, there was more of bacteria B than bacteria A at the start. B. Since y-intercepts indicate the amount of each bacteria at the start of the experiment, there was more of bacteria B than bacteria A at the start. C. Since the maximum value in the table for bacteria A is greater than the maximum value in the table for bacteria B, bacteria A has a faster growth rate than bacteria B. D. Since the minimum value in the table for bacteria A is less than the minimum value in the table for bacteria B, bacteria A has a slower growth rate than bacteria B.

Mathematics
1 answer:
Verizon [17]3 years ago
6 0

Answer:

A. False

B. True

C. False

D. False

Step-by-step explanation:

A. Since x-intercepts indicate the amount of each bacteria at the start of the experiment, there was more of bacteria B than bacteria A at the start.

False, it is the y-intercept of the function that indicates the amount at the start of the experiment.

B. Since y-intercepts indicate the amount of each bacteria at the start of the experiment, there was more of bacteria B than bacteria A at the start.

True, the y-intercept is given when x = 0, indicating the inicial value of the function.

C. Since the maximum value in the table for bacteria A is greater than the maximum value in the table for bacteria B, bacteria A has a faster growth rate than bacteria B.

False, because the maximum value of each table is given in different times, and also the inicial value of each table is different.

D. Since the minimum value in the table for bacteria A is less than the minimum value in the table for bacteria B, bacteria A has a slower growth rate than bacteria B.

False, the growth rate is not given by the inicial value. If we model both tables with an exponencial function, the count of bacteria A quadruped in two days, and the count of bacteria B doubled in one day, so they have the same growth rate.

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Geometry Helps please<br> (Only correct Answers only please) Thank you very much.
Levart [38]
<h3>Exact Answer: (160/3)pi</h3><h3>Approximate Answer: 167.47</h3>

The approximate answer results when you use pi = 3.14, and this approximate answer is rounded to two decimal places

note that (160/3)pi is the same as 160pi/3

========================================================

Work Shown:

r = radius = 4

h = height = 10

V = volume of cone

V = (1/3)pi*r^2*h

V = (1/3)pi*4^2*10

V = (160/3)pi exact answer

V = (160/3)*3.14

V = 167.46666....

V = 167.47 approximate answer rounded to two decimal places

8 0
3 years ago
Mark incorrectly evaluated the expression 4x+12-x to the power of 2 where is x-2. Identify and correct mark’s mistake.
nirvana33 [79]

Answer:

The value of 4x+12-x^2 is 16.

Step-by-step explanation:

The given expression is 4x+12-x^2\ at\ x=2.

Put x = 2 in the above expression as follows :

4x+12-x^2=4(2)+12-(2)^2\\\\=8+12-4\\\\=16

Hence, the final answer is 16.

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2 years ago
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Arte-miy333 [17]
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The greater of two numbers is one less than eight times the smaller . Their sum equals 98
Kitty [74]
B>a

b=8a-1

and we are told that 98=a+b so a=98-b, using this value for a in the equation above gives you:

b=8(98-b)-1

b=784-8b-1

b=783-8b

9b=783

b=87, and since a+b=98, a=11

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2 years ago
the figure below shows a circle centre of radius 10 cm the chord PQ=16cm calculate the area of the shaded region​
m_a_m_a [10]

Step-by-step explanation:

OPQ originally forms a sector, the formula for sector is

\frac{x}{360} \pi {r}^{2}

where x is the degrees of rotation between the two radii.

We know three sides length and is trying to find an angle between the radii so we can use law of cosines which states that

16 =  \sqrt{10 {}^{2} + 10 {}^{2}   - 2(100) \times  \cos(o) }

This isn't the standard formula, it's for this problem

16 =  \sqrt{200 - 200 \times  \cos(o) }

16 =  \sqrt{200  - 200 \cos(o) }

256 = 200 - 200 \cos(o)

56 =  - 200 \cos(o)

-  \frac{7}{25}  =  \cos(o)

\cos {}^{ - 1} (  - \frac{7}{25} )  =  \cos {}^{ - 1} ( \cos(o) )

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So we found our angle of rotation, which is 106.26

Now, we do the sector formula.

\frac{106.26}{360} (100)\pi

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is the area of sector.

Now let find the area of triangle, we can use Heron formula,

The area of a triangle is

\sqrt{s(s - a)(s - b)(s - c)}

where s is the semi-perimeter.

To find s, add all the side lengths up, 10,10,16 and divide it by 2.

Which is

\frac{36}{2}  = 18

\sqrt{18(18 - 10)(18 - 10)(18 - 16)}

\sqrt{18(8)(8)(2)}

\sqrt{(36)(64)}

6  \times 8 = 48

So our area of the triangle is 48. Now, to find the shaded area subtract the main area,( the sector of the circle) by the area of the triangle so we get

\frac{10626}{360}  - 48

Which is an approximate or

44.73

6 0
2 years ago
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