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3 years ago

11

Find a such that

1" title=" log_{3}(x) = alog_{5}(x) " alt=" log_{3}(x) = alog_{5}(x) " align="absmiddle" class="latex-formula">
for all values of x > 0 .

1 answer:

miss Akunina [59]3 years ago

3 0

Recall the change-of-base formula:

$\log_5x=\dfrac{\log_3x}{\log_35}$

(you can insert any logarithm on the right side)

It follows that $a=\log_35$ .

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4. A recipe for orange water says, "Mix 3 teaspoons yellow water with 1 teaspoon red water." For this recipe, we might say: "The

notsponge [240]

Answer:

2 batches of orange water = 6 tablespoon of yellow water and 2 tablespoon of red water

4 batches of orange water = 12 tablespoon of yellow water and 4 tablespoon of red water

Step-by-step explanation:

1 batch

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Red water = 1 teaspoons

Yellow water : Red water = 3 : 1

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3 years ago

Find the exact value of cos theta, given that sin thetaequalsStartFraction 15 Over 17 EndFraction and theta is in quadrant II.

vova2212 [387]

Answer:

$cos \theta = -\frac{8}{17}$

Step-by-step explanation:

For this case we know that:

$sin \theta = \frac{15}{17}$

And we want to find the value for $cos \theta$ , so then we can use the following basic identity:

$cos^2 \theta + sin^2 \theta =1$

And if we solve for $cos \theta$ we got:

$cos^2 \theta = 1- sin^2 \theta$

$cos \theta =\pm \sqrt{1-sin^2 \theta}$

And if we replace the value given we got:

$cos \theta =\pm \sqrt{1- (\frac{15}{17})^2}=\sqrt{\frac{64}{289}}=\frac{\sqrt{64}}{\sqrt{289}}=\frac{8}{17}$

For our case we know that the angle is on the II quadrant, and on this quadrant we know that the sine is positive but the cosine is negative so then the correct answer for this case would be:

$cos \theta = -\frac{8}{17}$

5 0

3 years ago