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ANTONII [103]
3 years ago
6

What is the degree of 5x^3+3x^2-4x+1? A.) 3 B.) 4 C.) 5 D.) 6

Mathematics
2 answers:
Aleks [24]3 years ago
7 0
When you say degree, do you mean answer if so, there is no solution to this equation.
ratelena [41]3 years ago
6 0

Answer:

Option (a) is correct.

The degree of  given polynomial  5x^3+3x^2-4x+1 is 3.

Step-by-step explanation:

Given : Polynomial 5x^3+3x^2-4x+1

We have to find the degree of the given polynomial  5x^3+3x^2-4x+1

Consider the given polynomial 5x^3+3x^2-4x+1

The degree of a polynomial is the highest power of the variable in the given polynomial.

Thus, for the given polynomial highest power is 3.

Thus, The degree of  given polynomial  5x^3+3x^2-4x+1 is 3.

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Tyler asks for a fish tank for his birthday. The task he wants is a rectangular prism 20 inches long, 10 inches wide, and 12 inc
Natasha_Volkova [10]

Answer:

2400 in^3

Step-by-step explanation:

Tyler asks for a fish tank for his birthday. The task he wants is a rectangular prism 20 inches long, 10 inches wide, and 12 inches tall. How much water will the fish tank hold?

the capacity of the tank can be determined by calculating the volume of the tank

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20 x 10 x 12 = 2400 in^3

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3 years ago
What is a polynomial​
luda_lava [24]

Answer:

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Step-by-step explanation:

6 0
3 years ago
9/20 divided by -3/5<br><br> Find the quotient.
hodyreva [135]
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6 0
2 years ago
Read 2 more answers
1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the
Musya8 [376]

Answer:

A) 34.13%

B)  15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

z=\frac{x-m}{s}

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

                                                =  0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587

C) Between 80 and 120

P( 80

D) less than 80

P( x < 80 ) = P( z

E) Between 70 and 100

P( 70

F) More than 130

P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013

8 0
3 years ago
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