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3 years ago

By graphing both sides of the equation, determine whether the following is an identity:

Mathematics

2 answers:

eimsori [14]3 years ago

4 0

Answer:

It is not an identity

Step-by-step explanation:

If you graphic the two equations (left and right) separately, if they are an identity, they will be the same graphic, which is not true in this case.

Another way in order to know if an equation is an identity, you can replace some values at x, for example 2 values:

X=30

X=60

And now we substitute in the equation, like this:

$1+sec^2(30)=tan^2(30)$

2,33=0,33 this is not equal on both sides

$1+sec^2(60)=tan^2(60)$

5=3 this is not equal on both sides

And if the results for each number x are the same on both sides of the equation, it is an identity. In this case they are different.

Alex3 years ago

3 0

Graphing is overkill... Let $x=0$ . Then $\sec0=1$ , while $\tan0=0$ . But $1+1=2\neq0$ , so this is not an identity.

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Which relations are functions? List the relation number(s) that are function(s) in the answer bank

Sergeu [11.5K]

Answer:

options 1,3,4 are functions.

Step-by-step explanation:

RULE: a relation is said to be a function if every element in the domain ( the numbers in the left side in the below sets) is related to only one number ( number on the right side in the below sets).

Let us check each option one by one:

1. 3 2

9 1

-4 7

0 -2

here each number on the left side is mapped to or is related to one number only.

so this relation is a function

2. 7 1

-5 2,3

1 0

here, "-5" is mapped to two different numbers. so this relation is not a function.

3. -2 -4

2 4

6 8

-6 -8

here each number on the left side is mapped to or is related to one number only.

so this relation is a function

4. 1 3

-1 3

2 3

-2 3

here each number on the left side is mapped to or is related to one number only.

so this relation is a function.

even if it is related to the same number, it doesn't matter.

it should follow the above given rule that's it.

6 0

3 years ago

!!Help!!<br>solve for x

Murrr4er [49]

WXY = XYW ( = 41 ) ⇒ XYW is a isosceles triangle

⇒ XW = YW

⇒ 54 = 6x + 6

⇒ 6( x + 1 ) = 54

⇒ x + 1 = 9

⇒ x = 8

ok done. Thank to me :>

4 0

3 years ago

Help pls pls pls gan i

ivanzaharov [21]

Answer:

C. 5 units

Step-by-step explanation:

look at the graph to the left

this seems too easy so I'm probably wrong but thats how I read the question

4 0

3 years ago

Chang buys a pack of 6 towels for $15.60.

sergeinik [125]

Answer:

$2.60 per towel

Step-by-step explanation:

15.60 devided by 6 =2.60

8 0

3 years ago

Read 2 more answers

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago