To answer this question, we are solving for the values of \(\displaystyle x\) that make this equation true.

To this, we need to get \(\displaystyle x\) on a side by itself so we can evaluate it. To do this, we first add \(\displaystyle 25x\) to both sides so that we can then begin to deal with the \(\displaystyle x^{2}\) value. So, for this data:

\(\displaystyle x^{2}-25x=0\rightarrow x^{2}=25x\)

\(\displaystyle x^{2}\)can also be written as \(\displaystyle x\cdot x\). Therefore:

\(\displaystyle x^{2}=25x\rightarrow x\cdot x=25x\)

Now we can divide both sides by \(\displaystyle x\) and find the value of \(\displaystyle x\).

\(\displaystyle \frac{x\cdot x}{x}=\frac{25x}{x}\rightarrow x=25\)

Therefore, the answer to this question is \(\displaystyle x=25\)

Line up each expression vertically. Then combine like terms to solve. 

\(\displaystyle 2x^3+6x^2-3x+1\)

\(\displaystyle x^3-2x^2+4x-3\)

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\(\displaystyle 2x^3+x^3 = 3x^3\)

\(\displaystyle 6x^2-2x^2 =4x^2\)

\(\displaystyle -3x+4x=x\)

\(\displaystyle 1-3=-2\)

Thus, the final solution is \(\displaystyle 3x^3+4x^2+x-2\).

\(\displaystyle 5x-5y=5y-5x\)

In adding \(\displaystyle 5x\) to both sides:

\(\displaystyle 5x-5y+5x=5y-5x+5x\)

\(\displaystyle 10x-5y=5y\)

. . .and adding \(\displaystyle 5y\) to both sides:

\(\displaystyle 10x-5y+5y=5y+5y\)

. . .the variables are isolated to become:

 \(\displaystyle 10x=10y\)

After dividing both sides by \(\displaystyle 10\), the equation becomes:

\(\displaystyle x=y\)

This is a problem where elimination can be help you save a little time. You can eliminate options quickly by simplifying one power at a time and comparing your work with the answer choices.

To begin, reorder the problem so that all like terms are next to each other. When doing so, keep an eye on your signs so that you don't accidentally make a mistake.

\(\displaystyle (5x^{3}-2x^{3}) + (31x^{2}+9x^{2}) +(-17x+34x) +(-6-12)\)

From here, combine each pair of terms. As you do so, compare your work with the answer choices.

\(\displaystyle (5x^{3}-2x^{3})=3x^{3} }\)   Eliminate any answer choices that have a different \(\displaystyle x^3\) term.

\(\displaystyle (31x^{2}+9x^{2})=40x^2\) Eliminate any answer choices that have a different \(\displaystyle x^2\) term.

\(\displaystyle (-17x+34x)=17x\) Eliminate any answer choices that have a different x term.

\(\displaystyle (-6-12)= -6+-12=-18\) Eliminate any answer choices that have a different constant term.

Once you put all of your solutions together, the correct answer looks like this:

\(\displaystyle 3x^{3} + 40x^{2} +17x -18\)

Using the binomial theorem, the term containing the x² y⁷ will be equal to

 (2x)²(–y)⁷

=36(–4x² y⁷)= -144x²y⁷

The easisest way to solve for \(\displaystyle a\) is to begin by plugging each pair of coordinates into the function.

Using our first point, we will plug in \(\displaystyle 0\) for \(\displaystyle x\) and \(\displaystyle 7\) for \(\displaystyle f(x)\).  This gives us the equation

\(\displaystyle 7=a(0)^2+b\).

Squaring 0 gives us 0, and multiplying this by \(\displaystyle a\) still gives 0, leaving only \(\displaystyle b\) on the right side, such that

\(\displaystyle 7 = b\).

We now know the value of \(\displaystyle b\), and we can use this to help us find \(\displaystyle a\).  Substituting our second set of coordinates into the function, we get

\(\displaystyle 19 = a(-2)^2+b\)

 which simplifies to

\(\displaystyle 19=4a+b\).

However, since we know \(\displaystyle b=7\), we can substitute to get

\(\displaystyle 19=4a+7\)

subtracting 7 from both sides gives

\(\displaystyle 12 = 4a\)

and dividing by 4 gives our answer

\(\displaystyle 3 = a\).

To answer this problem, we need to multiply the expressions together, being mindful of how to correctly multiply like variables with exponents. To do this, we add the exponents together if the the like variables are being multiplied and subtract the exponents if the variables are being divided. So, for the presented data:

\(\displaystyle 2x^{2} \cdot x^{3}y^{2} \cdot 3y = 2x^{5} \cdot 3y^{3}\)

We then multiply the remaining expressions together. When we do this, we will multiply the coefficients together and combine the different variables into the final expression. Therefore:

\(\displaystyle 2x^{5} \cdot 3y^{3} = 6x^{5}y^{3}\)

This means our answer is \(\displaystyle 6x^{5}y^{3}\).

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two \(\displaystyle x\) terms and one constant are multiplied; find the three products and add them, as follows:

\(\displaystyle \left (\underline{ x}+ 0.4\right ) (\underline{x} - 0.2) (3x\underline{-0.7})\)

\(\displaystyle x \cdot x \cdot (-0.7) = -0.7x^{2}\)

\(\displaystyle \left (\underline{ x}+ 0.4\right ) (x \underline{-0.2})(\underline{3x}-0.7)\)

\(\displaystyle x \cdot (-0.2) \cdot 3x= -0.6x^{2}\)

\(\displaystyle \left (x+ \underline{0.4}\right ) (\underline{x} - 0.2)(\underline{3x}-0.7)\)

\(\displaystyle 0.4 \cdot x \cdot 3x = 1.2 x^{2}\)

Add: \(\displaystyle -0.7x^{2}+ (-0.6x^{2})+ 1.2x^{2} = -0.1x^{2}\).

The correct response is \(\displaystyle -0.1\).

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two \(\displaystyle x\) terms and one constant are multiplied; find the three products and add them, as follows:

\(\displaystyle \left ( \underline{2x }+ \frac{1}{3}\right ) \left ( \underline{x}- \frac{1}{6}\right ) \left ( x\underline{+ \frac{1}{4}}\right )\)

\(\displaystyle 2x \cdot x \cdot \frac{1}{4} = \frac{1}{2}x^{2}\)

\(\displaystyle \left ( \underline{2x }+ \frac{1}{3}\right ) \left ( x \underline{- \frac{1}{6}}\right ) \left ( \underline{x}+ \frac{1}{4}\right )\)

\(\displaystyle 2x \cdot \left (- \frac{1}{6} \right ) \cdot x = - \frac{1}{3}x^{2}\)

\(\displaystyle \left ( 2x\underline{ + \frac{1}{3}}\right ) \left ( \underline{x}- \frac{1}{6}\right ) \left ( \underline{x}+ \frac{1}{4}\right )\)

\(\displaystyle \frac{1}{3} \cdot x \cdot x= \frac{1}{3}x^{2}\)

Add: \(\displaystyle \frac{1}{2}x^{2} +\left ( - \frac{1}{3}x^{2} \right ) + \frac{1}{3}x^{2} = \frac{1}{2}x^{2}\)

The correct response is \(\displaystyle \frac{1}{2}\).

If the expression \(\displaystyle \left ( A x + B\right )^{n}\) is expanded, then by the binomial theorem, the \(\displaystyle x^{k}\) term is

\(\displaystyle C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}\)

\(\displaystyle = C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}\)

or, equivalently, the coefficient of \(\displaystyle x^{k}\) is 

\(\displaystyle C(n, k) \cdot A ^{k} \cdot B ^{n - k}\)

Therefore, the \(\displaystyle x^{5}\) coefficient can be determined by setting 

\(\displaystyle A = 2, B =0.5, k=5, n = 8\):

\(\displaystyle C(8,5) \cdot 2^{5} \cdot 0.5 ^{8-5}\)

\(\displaystyle =C(8,5) \cdot 2^{5} \cdot 0.5 ^{3}\)

\(\displaystyle = 56\cdot 32 \cdot 0.125\)

\(\displaystyle =224\)