Writing in the form of power of 3

2 answers:

8 0

Writing in the form of power of 3 can be wrote a lot of ways. Any number to the power of 3 van also be read as the number to the power of 3, or the number cubed. Lets do some examples:-

3³ = 3 × 3 × 3 = 27
3³ = 27
3³ = three cubed = three to the power of 3

5³ = 5 × 5 × 5 = 125
5³ = 125
5³ = five cubed = five to the power of 3

(These are just 2 examples. There are MANY more.)
Hope I helped ya!!

fiasKO [112]3 years ago

3 0

Example : 3x3x3 Is one way you can write it Or you could do 3^3 were your second 3 is your exponent.
Hope this helps

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What are the potential solutions of log4x+log4(x+6)=2?

lions [1.4K]

The potential solutions of $log_4x+log_4(x+6)=2$ are 2 and -8.

<h3>Properties of Logarithms</h3>

From the properties of logarithms, you can rewrite logarithmic expressions.

The main properties are:

Product Rule for Logarithms - $log_{b}(a*c)=log_{b}a+log_{b}c$
Quotient Rule for Logarithms - $log_{b}(\frac{a}{c} )=log_{b}a-log_{b}c$
Power Rule for Logarithms - $log_{b}(a^c)=c*log_{b}a$

The exercise asks the potential solutions for $log_4x+log_4(x+6)=2$ . In this expression you can apply the Product Rule for Logarithms.

$log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0$

Now you should solve the quadratic equation.

Δ= $b^2-4ac=36-4*1*(-16)=36+64=100$ . Thus, x will be $x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}$ . Then:

$x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8$

The potential solutions are 2 and -8.

Read more about the properties of logarithms here:

brainly.com/question/14868849

4 0

2 years ago

What is the equation of the quadratic graph with a focus of (-4,17/8) and a directrix of y=15/8

Temka [501]

So hmm notice the graph below

based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"

$\bf \begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})}\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}$

now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be $\bf \cfrac{17}{8}-\cfrac{15}{8}=\cfrac{2}{8}\iff \cfrac{1}{4}$

well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either

what's the "p" distance? well, we just found it, is just 1/8

so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8

and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic