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3 years ago

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What is the maximum number of terms a sixth-degree polynomial function in standard form can have? I have this question and I hav

e no idea what's the answer.

1 answer:

LenaWriter [7]3 years ago

8 0

Answer:

The maximum number of terms is 7

Step-by-step explanation:

A sixth degree polynomial has the highest polynomial power raised to the power of 6

It can be written in the form;

6x^6 + 5x^5 + 4x^4 + 3x^3 + 2x^2 + x + 1

So the maximum number of terms is;

7

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Triphasil-28 birth control tablets are taken sequentially, 1 tablet per day for 28 days, with the tablets containing the followi

bazaltina [42]

Answer:

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

Step-by-step explanation:

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3}$

In which $L_{1}, L_{2}$ and $L_{3}$ are the number of miligrams of levonorgestrel taken in each phase.

The total milligrams of ethinyl estradiol is:

$E = E_{1} + E_{2} + E_{3}$

In which $E_{1},E_{2}$ and $E_{3}$ are the number of miligrams of ethinyl estradiol taken in each phase.

We can solve this by phase.

Phase 1: 6 tablets, each containing 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$6*0.050mg = 0.30mg$ of levonorgestrel and $6*0.030mg = 0.18mg$ of ethinyl estradiol are taken.

So $L_{1} = 0.30$ and $E_{1} = 0.18$

Phase 2: 5 tablets, each containing 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol

In this phase,

$5*0.075mg = 0.375mg$ of levonorgestrel and $5*0.040mg = 0.20mg$ of ethinyl estradiol are taken.

So $L_{2} = 0.375$ and $E_{2} = 0.20$

Phase 3: 10 tablets, each containing 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$10*0.125mg = 1.25mg$ of levonorgestrel and $10*0.030mg = 0.30mg$ of ethinyl estradiol are taken.

So $L_{3} = 1.25$ and $E_{3} = 0.30$

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3} = 0.30 + 0.375 + 1.25 = 1.925$

The total milligrams of ethinyl estradiol is:

$E = E_{1} + E_{2} + E_{3} = 0.18 + 0.20 + 0.30 = 0.68$

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

6 0

3 years ago

mixas84 [53]

-3x = -3 because you would multiply one of the equations by -1 to eliminate y.

5 0

3 years ago

Simplify to create an equivalent expression. 1+4(6p−9)

myrzilka [38]

Answer: -35+24p

Step by step in picture below

3 0

3 years ago

The wheels on Noah's bike have a circumference of about 5 feet.

ANTONII [103]

Answer:

75 feet

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A researcher is going to estimate the average typing speed of students of a college. He selects a random sample of 20 students a

Stella [2.4K]

Answer:

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 1.7291

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have $s = 5, n = 20$ . So

$M = T\frac{s}{\sqrt{n}}$

$M = 1.7291\frac{5}{\sqrt{20}}$

$M = 1.933$

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

8 0

3 years ago