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lana [24]
3 years ago
11

Describe one difference between rewriting a literal equation and solving an equation in one variable.

Mathematics
1 answer:
sesenic [268]3 years ago
3 0

Literal equation:

we know that

literal equation is equation of more one variable

for example: 2x+3y=1

And we can solve such equations only if we are given two such equation

because it has two variables

One variable equation:

such equation has single variables

for example:

3x=21

we don't need any other equation

we can solve for variables from single equation itself

because it has only one variable that is x

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7,4,9,8,4,4,6,6 mean,mode,median, and range
skad [1K]

Answer:

Mean: 6    Mode: 4   Median: 6    Range:5

Brainliest please :)

5 0
2 years ago
Read 2 more answers
(5x3 + 3)2 = (5x3)2 + (3)2 = 25x6 + 9 is this correct
faust18 [17]

Answer:

No!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! NO NO NO NO NO... SEE BELOW!!!!!!

Step-by-step explanation:

LOL I like caps!!! and !'s

To question

If 5x3 is 5x^3 then...

5x^3 times 2 equals 10x^3 no 10x^6

So the end result will be 10x^3\\ + 9

5 0
3 years ago
Hi, what is this number in expanded form, please let me know?
Nutka1998 [239]

Answer:

<em>5,078.401 </em>

Step-by-step explanation:

5,000 + 70 + 8 + 0.4 + 0.001 = <em>5,078.401</em>

7 0
3 years ago
In a test gender selection technique results consisted of 250 baby girls and 234 boys. Based on this result what is the probabil
Assoli18 [71]

Answer: 125/242

Step-by-step explanation: Probability is favorable outcomes over all outcome, since there were 250+234 total outcome, and 250 were favorable (girls), then the probability is 250/250+234=250/484=125/242

8 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
astra-53 [7]

Answer:

3\pi \rightarrow y=2\cos \dfrac{2x}{3}\\ \\\dfrac{2\pi }{3}\rightarrow y=6\sin 3x\\ \\\dfrac{\pi }{3}\rightarrow  y=-3\tan 3x\\ \\10\pi \rightarrow y=-\dfrac{2}{3}\sec \dfrac{x}{5}

Step-by-step explanation:

The period of the functions y=a\cos(bx+c) , y=a\sin(bx+c), y=a\sec (bx+c) or y=a\csc(bx+c) can be calculated as

T=\dfrac{2\pi}{b}

The period of the functions y=a\tan(bx+c) or y=a\cot(bx+c) can be calculated as

T=\dfrac{\pi}{b}

A. The period of the function y=-3\tan 3x is

T=\dfrac{\pi}{3}

B. The period of the function y=6\sin 3x is

T=\dfrac{2\pi}{3}

C. The period of the function y=-4\cot \dfrac{x}{4} is

T=\dfrac{\pi}{\frac{1}{4}}=4\pi

D. The period of the function y=2\cos \dfrac{2x}{3} is

T=\dfrac{2\pi}{\frac{2}{3}}=3\pi

E. The period of the function y=-\dfrac{2}{3}\sec \dfrac{x}{5} is

T=\dfrac{2\pi}{\frac{1}{5}}=10\pi

5 0
3 years ago
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