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3 years ago

Which of the following expression(s) are fourth-degree trinomials? Select all that apply.

Mathematics

1 answer:

frozen [14]3 years ago

6 0

Answer:

Options A and B are polynomial of the fourth degree

Step-by-step explanation:

In this problem, option

A. 3x2y + 5x3y + 6y4

Is a Polynomial of the fourth degree because of the 6y⁴ term which is the highest degree

Also the option

B. 6y4 + 5x3 + 1 has a 6y⁴ term which indicates that the polynomial is a fourth degree polynomial

What is the degree of a polynomial?

In mathematics, the degree of a polynomial is the highest of the degrees of the polynomial's monomials with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer

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Using the Laws of exponents, simplify the following problem.

den301095 [7]

$\frac{4 {}^{4} }{ {4}^{ - 8} ({4}^{ - 14}) } = \frac{4 {}^{4} }{ {4}^{ - 22 } }$

(4^4)/(4^-22)
4^(4-(-22))
4^26

Follow my Account.

7 0

3 years ago

Which of the following best describes how to solve<br> 2x+3 = 2x + 37

weeeeeb [17]

Answer: The first option.

Step-by-step explanation:

If you square you get $2x+3=2x+9$

x=0

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3 years ago

Sonya put $400 in a savings account for 3 years which pays 5% annual simple interest. The formula to determine the amount of int

Aleks04 [339]

Answer:

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3 years ago

manuel deposits $10000 for 12 yr in an account paying 4% compounded annually.He then puts this total amount on deposit in anothe

Naddik [55]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$10000\\
r=rate\to 4\%\to \frac{4}{100}\to &0.04\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &12
\end{cases}
\\\\\\
A=10000\left(1+\frac{0.04}{1}\right)^{1\cdot 12}\implies A=1000(1.04)^{12}\\\\\\ A\approx 16010.32$

he then turns around and grabs that money and sticks it for another 9 years,

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
~~
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$16010.32\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\
t=years\to &9
\end{cases}
\\\\\\
A=16010.32\left(1+\frac{0.05}{2}\right)^{2\cdot 9}\implies A=16010.32(1.025)^{18}
\\\\\\
A\approx 24970.64$

add both amounts, and that's how much is for the whole 21 years.

6 0

4 years ago

7.49362 to the nearest no thousandth

True [87]

7.494 is the nearest thousandth

3 0

3 years ago

Read 2 more answers