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yaroslaw [1]
3 years ago
5

If f(x) is a linear function and the domain of f(x) is the set of all real numbers, which statement cannot be true?

Mathematics
1 answer:
IRISSAK [1]3 years ago
3 0

Answer:

can u please add the statements.

Step-by-step explanation:

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Solving the equation
creativ13 [48]

Answer:

9s+2=9s-5

Step-by-step explanation:

2 times 9s and 2 times 2 then subtract your 9sequals 3 times 3sand 3 times 6 so then you get 9s and 18 then you subtract 23 and you get 23. then you get your answer 9s-5=9s+2

5 0
3 years ago
NEED HELP WORTH 100 POINTS SHOW YOUR WORK PLEASE
tia_tia [17]

Answer:

Q1:

-5\sqrt{27c^2}

-5\sqrt{27}\sqrt{c^2}

-5\times\sqrt{9} \sqrt{3} \sqrt{c^2}

-5\times3\sqrt{3}\sqrt{c^2}

-5\times3\sqrt{3}c

-15\sqrt{3}c

Q2:

5\sqrt{72}-3\sqrt{32}

5\sqrt{36} \sqrt{2} -3\sqrt{16} \sqrt{2}

5\times6\sqrt{2}-3\times4\sqrt{2}

30\sqrt{2}-12\sqrt{2}

18\sqrt{2}

Q3:

\sqrt{108yz}+3\sqrt{98yz}+2\sqrt{75yz}

\sqrt{36} \sqrt{3} \sqrt{yz} +3\sqrt{49} \sqrt{2} \sqrt{yz} +2\sqrt{25} \sqrt{3} \sqrt{yz}

6\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}+10\sqrt{3}\sqrt{yz}

16\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}

Q4:

\sqrt{15n^2}\sqrt{10n^3}

\sqrt{15}n\sqrt{10n^3}

\sqrt{15}n\sqrt{10}\sqrt{n^2}\sqrt{n}

\sqrt{15}\sqrt{10}n^2\sqrt{n}

\sqrt{5}\sqrt{3}\sqrt{5}\sqrt{2}n^2\sqrt{n}

5\sqrt{3}\sqrt{2}n^2\sqrt{n}

5\sqrt{6}n^2\sqrt{n}

Q5:

\frac{\sqrt{3x^2y^3}}{4\sqrt{5xy^3}}

\frac{\sqrt{3}xy\sqrt{y}}{4\sqrt{5}y\sqrt{x}\sqrt{y}}

Cancel off \sqrt{x}:

\frac{\sqrt{3}y\sqrt{x}\sqrt{y}}{4\sqrt{5}y\sqrt{y}}

Cancel off y:

\frac{\sqrt{3}\sqrt{x}\sqrt{y}}{4\sqrt{5}\sqrt{y}}

Cancel off \sqrt{y} :

\frac{\sqrt{3}\sqrt{x}}{4\sqrt{5}}

Multiply \sqrt{5} on the numerator and denominator:

\frac{\sqrt{3}\sqrt{x}\sqrt{5}}{4\sqrt{5}\sqrt{5}}

\frac{\sqrt{15}\sqrt{x}}{20}

5 0
3 years ago
Greek city-states began to form in 800 B.C. and in 200 A.D, Christians began getting persecuted. How many years passed between t
asambeis [7]

Answer: 1000 years

Step-by-step explanation:

-Greek city-states began to form in 800 B.C i.e 800 years Before Christ appearance

- up till 200 A.D i.e 200 years of After Christ disappearance

Hence, the number of years between the time that Greek city-states began to form and the persecution of Christians began is equal to the sum of years Greek city States started formation and Year persecution began

i.e 800 BC + 200 A.D = 1000 Years

Thus, the number of years of that passed was 1000 years

6 0
3 years ago
See image attached below keeeeeeeeeeeeeeeeeeeed
emmasim [6.3K]

Answer:

.13%

68.26%

2.28%

47.72%

49.87%

34.13%

Step-by-step explanation:

1.) standardize by subtracting the mean and dividng by the standard deviation

(625-1000)/125= -3

go to a ztable to get .0013 or (1-.9987)

this is equal to .13%

2.)

875<x<1125

standardize both seperately and subtract them

(875-1000)/125= -1  whose probability is .1587 (or 1-.8413)

(1125-1000)/125= 1  whose probability is .8413

.8413-.1587= 68.26%

3.)

Find the probability that someone paid less than 1250 and take its compliment

(1250-1000)/125= 2 which has a probability of .9772

take its compliment (1-.9772)= .0228= 2.28%

4.)

Same process as question 2

(750-1000)/125= -2 which has a probability of (1-.9772) = .0228

(1000-1000)/125= 0 which has a probability of .5

.5-.0228= .4772= 47.72%

5.) same deal as the previous question

(625-1000)/125= -3 which has a probability of (1-.9987)= .0013

(1000-1000)/125= 0 which has a probability of .5

.5-.0013= .4987= 49.87%

6.)same deal the previous question

(875-1000)/125= -1 which has a probability of (1-.8413)=.1587

(1000-1000)/125= 0 which has a probability of .5

.5-.1587= .3413= 34.13%

4 0
3 years ago
How can you use expressions and models to determine if expressions are equivalent
Hunter-Best [27]

To check for equivalence, use properties of operations such as the Distributive Property. Two expression are said to be equivalent if they resulted in the same number after evaluation of each. You can both expand and factor expressions to generate equivalent expressions. 

6 0
3 years ago
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