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1 year ago

PLS HELP!!

Mathematics

1 answer:

inna [77]1 year ago

6 0

The correct answer are as follows:

A. False

B. True

C. False

D. False

<h3>What is Function?</h3>

The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input x.

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 initial 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 initial 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 initial value. If we model both tables with an exponential 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.

Learn more about Function from:

brainly.com/question/12431044

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Determine whether line segment FG is tangent to circle E. <br><br> True or False

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which is not true.
Therefore, line segment FG is not tangent to circle E.

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2 years ago

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malfutka [58]

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3 years ago

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What is the solutions to 3x+y=-8 and -2x+y=6

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3 years ago

Use complete sentences to describe why Set A = { X | X is an even whole number between 0 and 2} = ∅

RoseWind [281]

Consider the Set A = {X | X is an even whole number between 0 and 2 } = $\Phi$ .

Since, whole numbers are the set of numbers starting from zero upto infinity.

Even numbers are the numbers which are exactly divisible by '2'.

So, we have to find the even whole number between 0 and 2.

Since, only '1' is a whole number between 0 and 2 which is not an even number as '1' is not divisible by '2'.

Therefore, there is no even whole number between 0 and 2.

So, this set is empty.

Therefore, A = { X | X is an even whole number between 0 and 2} = $\Phi$

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2 years ago

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago