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2 years ago

Name the type of polynomial indicating the

Mathematics

1 answer:

lutik1710 [3]2 years ago

5 0

first off, bear in mind that the terms are usually separated by a + or a - sign.

secondly, the degree of a term of a polynomial comes from the sum of its variables' exponents.

$\bf 2x^3y^2\qquad \begin{cases} \textit{no + or - signs, just one term a \underline{monomial}}\\\\ x^3y^2\implies \stackrel{degree}{3+2}\implies 5 \end{cases}$

$\bf \stackrel{term}{5x}+\stackrel{term}{3x^2}-\stackrel{term}{9}\qquad \begin{cases} \textit{three terms, a \underline{trinomial}}\\\\ 5x^1\implies 1\\ 3x^2\implies 2\qquad \stackrel{\textit{degree of the polynomial}}{\textit{term with highest degree}}\\ \end{cases}$

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What’s the length of AZ. If A is between Y and Z, Y A = 22, AZ = 16x, and Y Z = 166

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Answer:

c) 144

Step-by-step explanation:

Since A is between Y and Z,

$YZ = YA + AZ$

$116 = 22 + 16x\\16x = 116 - 22\\16x = 144\\x = 9$

however, the question asks for $the length of AZ$ , not the value of x.

Thus,

$AZ = 16x$

$AZ = 16\\$ × $9$

$AZ = 144$

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2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

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Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

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Solve for $u$ :

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$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

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