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2 years ago

Which powers have of value of 81. Possible Answers: 1^81, 3^4, 6^4, 9^2, 27^3.

Mathematics

2 answers:

pishuonlain [190]2 years ago

7 0

1^81 is 1 multiplied by itself 81 times, which equals 1

3^4 is 3 multiplied by itself 4 times (3 x 3 x 3 x 3), which equals 81

6^4 is 6 multiplied by itself 4 times (6 x 6 x 6 x 6), which is 1296

9^2 is 9 multiplied by itself twice (9 x 9), which is 81

27^3 is 27 multiplied by itself 3 times (27 x 27 x 27), which is 19683.

Answer: 3^4 and 9^2

quester [9]2 years ago

4 0

1^81, 9^2, 3^4. the others are rather to high or to low

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Answer:

The correct answer is A

Step-by-step explanation:

The secant is the distance from the center to the vertical tangent in a certain direction.

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2 years ago

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2 years ago

Ms. Houston's fifth grade class is going to A museum the class raises $68 for the trip transportation to the museum cost $40The

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3 years ago

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-2.1b +5.3= b -- 3.1b + 5.3<br><br> does is have one solution or infinitely many solutions ? Why ?

pychu [463]

Answer: infinitely many solutions.

Step-by-step explanation:

Ok, our equation is:

-2.1*b + 5.3 = b - 3.1*b + 5.3

now, simplifyng the right side, we have:

b - 3.1*b + 5.3 = (1 - 3.1)*b + 5.3 = -2.1*b + 5.3

Then our initial expression is:

-2.1*b + 5.3 = -2.1*b + 5.3

So in both sides of the equality we have the exact same thing, so this is a trivial equality.

This means that the equality will remain true for any value of b, which means that we have infinitely many solutions.

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2 years ago

PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

Explanation:

The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$

Thus, both sides of the equation are equal.

Hence, $x=\frac{5}{2}$ is a true solution.

Thus, solutions are not extraneous.

Hence, Option C is the correct answer.

8 0

3 years ago