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3 years ago

What is log15(2)^3 rewritten using the power property?

2 answers:

6 0

Power property:
$log_{a} b^m=m \cdot log_a b$

That means the power m can be brought down to front of the logarithm.
Using the power property, we get
$log_{15}2^3=3 \cdot log_{15}2$

NNADVOKAT [17]3 years ago

6 0

Answer:

D. 3log15^2

Step-by-step explanation:

I just took the test its right! give me my props! <3

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In ΔFGH, the measure of ∠H=90°, GF = 53, HG = 28, and FH = 45. What ratio represents the sine of ∠G?

andrew11 [14]

we have been given that in ΔFGH, the measure of ∠H=90°, GF = 53, HG = 28, and FH = 45. We are asked to find the ratio that represents the sine of ∠G.

First of all, we will draw a right triangle using our given information.

We know that sine relates opposite side of right triangle with hypotenuse.

$\text{sin}=\frac{\text{Opposite}}{\text{hypotenuse}}$

We can see from the attachment that opposite side to angle G is FH and hypotenuse is GF.

$\text{sin}(\angle G)=\frac{FH}{GF}$

$\text{sin}(\angle G)=\frac{45}{53}$

Therefore, the ratio $\frac{45}{53}$ represents the sine of ∠G.

7 0

4 years ago

(cos x - (sqrt 2)/2)(sec x -1)=0

Darya [45]

$(\cos x-\frac{\sqrt{2}}{2})(\sec x-1)$ =0 [/tex]

$=(\cos x-\frac{1}{\sqrt{2}})(\sec x-1)$ =0 [/tex]

$\frac{(\sqrt{2}\cos x-1)}{\sqrt{2}}(\frac{1}{\cos x\ }-1)=0$

(Reciprocal Identity)

$(\frac{^{\sqrt{2}\\cos x-1}}{^{\sqrt{2}}})(\frac{1-\cos x}{\cos x})=0$

$\frac{^{(\sqrt{2}\cos x-1})}{\sqrt{2}}\frac{(1-\cos x)}{\cos x}=0$

$(\sqrt{2}\cos x-1}){(1-\cos x)}=0$ (ZeroProduct Property)

$\sqrt{2}\cos x-1=0$

$\sqrt{2}\cos x=1$

$\cos x=\frac{1}{\sqrt{2}}$

$x=\frac{\Pi }{4}$

and

$1-\cos x=0$

$\cos x=1$

$x=0$

x=0 and x= $\frac{\Pi }{4}$ are the solutions.

6 0

4 years ago

Read 2 more answers

Evaluate the integral by interpreting it in terms of areas Draw a picture of the region the integral

denis23 [38]

Answer:

Step-by-step explanation:

The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.

For the integral from [0, 2], the equation of the line is -3x + 6;

For the integral from [2, 3], the equation of the line is 3x - 6.

We integrate then:

$\int\limits^2_0 {-3x+6} \, dx+\int\limits^3_2 {3x-6} \, dx$ and

$-\frac{3x^2}{2}+6x\left \} {{2} \atop {0}} \right. +\frac{3x^2}{2}-6x\left \} {{3} \atop {2}} \right.$ sorry for the odd representation; that's as good as it gets here!