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2 years ago

Explain how to evaluate 3 to the fourth power

Mathematics

1 answer:

Nikolay [14]2 years ago

5 0

Answer:

3 to the fourth power would basically mean 3 x 3 x 3 x 3

Step-by-step explanation:

Alot of people get this confused with 3 x 4 but when multuply to the power of four its 3 x 3 x 3 x 3 which is 81

Brainliest?

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I can’t figure this out!! please helppp

krok68 [10]

Answer:

The third answer

Step-by-step explanation:

(side note: You probably need to use a calculator that can calculate pi)

The formula to find the area of a circle is pi r squared.

In this case, pi times r squared is 19.63495....... continued

This is probably going to be rounded

so... 19.63

For all of the circles, this is multiplied by 4.

This is 78.52

The area of the square is 100 cm since 10 times 10 is 100

You take 100 cm and then subtract it by 78.54

Voila! now you have 21.46

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2 years ago

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noname [10]

The right answer is 4

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2 years ago

What is the rotation?

Vlad1618 [11]

It means to move something

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2 years ago

Zahra plans to draw a triangle with angles measuring 50, 60, and 70. Which is true about Zahra’s plan?

andriy [413]

<span>D. Zahra can make more than one triangle with these angle measures.
(Infinite similar triangles)</span>

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2 years ago

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Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (1 point)

sesenic [268]

Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3> or , $u = -5i-4j , v=-4i-3j$

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

$u.v =|u|(|v|)cosx$ , where x is angle between u & v !

⇒ $u.v =|u|(|v|)cosx$

⇒ $cosx =\frac{u.v}{|u|(|v|)}$

Now , $u.v = (-5i-4j)(-4i-3j)$

⇒ $u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)$

⇒ $u.v = 20+12$ { $i(j) = j(i) =0$ }

⇒ $u.v = 32$

Now , Modulus of any vector $r = xi+yj$ is $|r| = \sqrt{x^{2}+y^{2}}$ So ,

$|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$

Putting all these values in equation $cosx =\frac{u.v}{|u|(|v|)}$ we get:

⇒ $cosx =\frac{32}{5(\sqrt{41})}$

⇒ $cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})$

⇒ $x =cos^{-1}(1)$ { $cos0 = 1$ }

⇒ $x =0$ °

Therefore , Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

8 0

2 years ago