For the answer to the question above, the power of "y" doesn't influence descending order for x even if it is higher than power of x.

Unordered polynomial is
7x3y3 + 4 - 11x5y2 - 3x2y

Polynomial in descending order looks like this
-11x5y2 + 7x3y3 - 3x2y + 4

Degree of the second term is 5 ( because of x5 )

3 0

3 years ago

Read 2 more answers

Help me please<br>help.me

levacccp [35]

Answer:

The answers of the questions are given below :

a) = 4096
b) = 1.25
3) = m²
4) = r⁴s³
5) = a⁸/b¹²

Step-by-step explanation:

$\large{\tt{\underline{\underline{\red{QUESTION}}}}}$

$\begin{gathered}\footnotesize\boxed{\begin{array}{c|c|c}\bf\underline{Given}&\bf\underline{Solution}&\bf{\underline{Simple\: Form}}\\\\\rule{60pt}{0.5pt} &\rule{70pt}{0.5pt}& \rule{70pt}{0.5pt}\\\\ 1.\: {4}^{6} & & \\\\ 2.\: \bigg(\dfrac{2^6}{5^3} \bigg)& &\\\\ 3. \: \Big({m}^{\frac{2}{3}}\Big)\bull \Big({m}^{\frac{4}{3}}\Big) & &\\\\4. \: \big({r}^{12} {s}^{9}\big)^{ - \frac{1}{3}} &&\\\\ 5.\bigg(\dfrac{a^4}{a^6}\bigg)^{2}& &\end{array}}\end{gathered}$

$\begin{gathered}\end{gathered}$

$\large{\tt{\underline{\underline{\red{SOLUTION}}}}}$

Question. 1

>> 4⁶

$\begin{gathered}\qquad{= 4 \times 4 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 16 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 64 \times 4 \times 4 \times 4} \\ \qquad{= 256\times 4 \times 4} \\ \qquad{= 1024 \times 4} \\ \qquad{= 4096} \end{gathered}$

Hence, the answer is 4096.

$\begin{gathered}\end{gathered}$

Question. 2

>> (2⁶/5³)^-⅓

$\begin{gathered} \qquad\implies{\bigg(\frac{2^6}{5^3}\bigg)^{ - \frac{1}{3}}}\\ \\ \qquad\implies{\bigg(\frac{64}{125}\bigg)^{ - \frac{1}{3}}}\\ \\\qquad\implies{\bigg( \frac{1}{\frac{64}{125}}\bigg)^{ \frac{1}{3}}} \\ \\ \qquad\implies{\bigg( 1 \times \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\ \\ \qquad\implies{\bigg( \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\ \\\qquad\implies{\bigg( \sqrt[3]{ \frac{125}{64}}\bigg)} \\ \\ \qquad\implies{\bigg( \dfrac{5}{4} \bigg)} \\ \\ \qquad\implies{\Big( 1.25\Big)}\end{gathered}$

Hence, the answer is 1.25.

$\begin{gathered}\end{gathered}$

Question. 3

>> (m^2/3)•(m^4/3)

$\begin{gathered} \qquad{= \Big({m}^{\frac{2}{3}}\Big) \bull \Big({m}^{ \frac{4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{2}{3} + \frac{4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{2 + 4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{6}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{2}\Big)}\end{gathered}$

Hence, the answer is m².

$\begin{gathered}\end{gathered}$

Question. 4

>> (r¹² s⁹)^⅓

$\begin{gathered} \qquad\implies{\Big( {r}^{12} \: {s}^{9}\Big)^{\frac{1}{3}}}\\\\ \qquad\implies{\Big({r}^{\frac{12}{3} } \: {s}^{\frac{9}{3}}\Big)} \\ \\ \qquad\implies{\Big({r}^{\cancel{\frac{12}{3}}} \: {s}^{\cancel{\frac{9}{3}}}\Big)} \\ \\ \qquad\implies{\Big({r}^{4} \: {s}^{3}\Big)} \end{gathered}$

Hence, the answer is r⁴s³.

$\begin{gathered}\end{gathered}$

Question. 5

>> (a⁴/b⁶)^2

$\begin{gathered} \qquad{ = \Big(\frac{a^4}{b^6}\Big)^{2}} \\ \\ \qquad{ = \Big(\frac{a^{4 \times 2}}{b^{6 \times 2}}\Big)} \\ \\ \qquad{ = \Big(\frac{a^{8}}{b^{12}}\Big)} \end{gathered}$

Hence, the answer is a⁸/b¹².

$\underline{\rule{220pt}{3pt}}$

4 0

2 years ago

Help me with this. I am really stuck.

Kryger [21]

By the Hypotenuse Leg theorem, the two triangles are congruent. That is, ΔBXA ≅ ΔBYA

From the question, we are to prove that ΔBXA ≅ ΔBYA

From the Hypotenuse Leg theorem which states that "two right triangles are congruent if the hypotenuse and one leg of one right triangle are congruent to the other right triangle's hypotenuse and leg side".

From the given information,

BX ⊥ XA.

ΔBXA is right triangle with hypotenuse AB

Also,

BY ⊥ YA.

ΔBYA is right triangle with hypotenuse AB

∴ The hypotenuses of the triangles are congruent

Also, from the given information,

XA is congruent to YA

XA and YA are the legs of the right triangles.

Hence, by the Hypotenuse Leg theorem, the two triangles are congruent. That is, ΔBXA ≅ ΔBYA

Learn more on Congruent triangles here: brainly.com/question/27983954

#SPJ1

7 0

1 year ago