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2 years ago

Number 2 says Write the given equation in exponential form. Log8 1/512=-3

Mathematics

1 answer:

MA_775_DIABLO [31]2 years ago

7 0

In exponential form it is given as, 8⁻³ = $$\frac{1}{512}$

Step-by-step explanation:

We have to write the given equation in the exponential form, that is in the form of base to the power value.

Given:

log₈ ( $$\frac{1}{512}$ ) = -3

Here the base is 8.

In the exponential form, it is written as,

base to the power 3 = 512

Here the reciprocal of 512 is given, so,

base to the power -3 = $$\frac{1}{512}$

So the exponential form is given as, 8⁻³ = $$\frac{1}{512}$

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I will mark you brainiest if you can answer this

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What is the sum of 8/10 and 8/100

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8/10 = 80/100
So the sum is 80/100 + 8/100 = 88/100 or 22/25

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3 years ago

If tan theta = 10/13 and cos theta > 0 then sin2theta is what ?

svetoff [14.1K]

$sin2\alpha = \frac{260}{269}$

Step-by-step explanation:

We have , $Tan\alpha = \frac{Perpendicular}{Base} = \frac{10}{13}$ ,

We know that $sin\alpha = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }$

Substituting values of P & B , $sin\alpha = \frac{10}{\sqrt{10^{2} + 13^{2}} } = \frac{10}{\sqrt{269} }$

Now , $sin2\alpha = 2sin\alpha cos\alpha = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }$

⇒ $sin2\alpha =$ $\frac{10}{\sqrt{269} }$ × $\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }$ ×2

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⇒ $sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} } )$

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3 years ago

What number do you multiply 5/3 by to get the product of 1

Usimov [2.4K]

Answer:

3/5

Step-by-step explanation:

Any number multiplied by it's reciprocal will get you the product of 1

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3 years ago

Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims tha

GalinKa [24]

Answer:

$\theta_1 \ n\ \theta_2 = 120, 240$

Step-by-step explanation:

The question is incomplete, as the angles of rotation are not stated.

However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself

We have:

$Nonagon = 9\ sides$

$Hexagon = 6\ sides$

Divide 360 degrees by the number of sides in each angle, then find the multiples.

Nonagon

$\theta = \frac{360}{9} =40$

List the multiples of 40

$\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320$

Hexagon

$\theta = \frac{360}{6} =60$

List the multiples of 60

$\theta_2 = 60, 120, 180, 240, 300$

List out the common angles

$\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320$

$\theta_2 = 60, 120, 180, 240, 300$

$\theta_1 \ n\ \theta_2 = 120, 240$

This means that, only a rotation of $120, 240$ will lift both shapes onto themselves, when applied to both shapes.

The other angles will only work on one of the shapes, but not both at the same time.

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3 years ago