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3 years ago

I honestly don't know what to put here. Attachment below.

Mathematics

1 answer:

irakobra [83]3 years ago

4 0

Answer: $(5^{-2} )^{2}$ and $5^{2} *5^{-6}$

Step-by-step explanation:

$\frac{1}{5} *\frac{1}{5} *\frac{1}{5} *\frac{1}{5} =\frac{1}{625}$ , you can now see which of these equal 1/625.

$(5^{-2} )^{2}=\frac{1}{625}$

$(5^{-4})^{0}\neq \frac{1}{625}$

$\frac{5^{1} }{5^{4}} \neq \frac{1}{625}$

$5^{2} *5^{-6}= \frac{1}{625}$

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)

- Cutiepatutie ☺❀❤

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Step-by-step explanation:

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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Answer: The correct option is (B).

True. The vector V₃ is a linear combination of V₁ and V₂, so at least one of the vectors in the set is a linear combination of the others and set is linearly dependent.

Step-by-step explanation:

Given that;

If V₁ ....... V₄ are in R⁴ and V₃ = 2V₁ + V₂ then {V₁, V₂, V₃, V₄} is linearly dependent.

Lets {V₁, V₂, V₃, V₄} are linearly dependent

Then there exist a scalars C₁, C₂, C₃, C₄

So that C₁V₁ + C₂V₂ + C₃V₃ + C₄V₄ = 0

where at least one of the Ci ≠ 0.

Take C₃ ≠ 0 then we have V₃ = (C₁V₁ + C₂V₂ + C₄V₄) / C₃

V₃ is a linear combination of V₁, V₂ and V₄}

that is

Given V₃ = 2V₁ + V₂

⇒ 2V₁ + V₂ - V₃ = 0

⇒ 2V₁ + V₂ + V₃ + 0V₃ = 0

Here, C₁ = 1, C₂ = 1, C₃ = -1 and C₄ = 0

So that {V₁, V₂, V₃, V₄} is linearly dependent.

therefore option B id the right answer.

- True. The vector V₃ is a linear combination of V₁ and V₂, so at least one of the vectors in the set is a linear combination of the others and set is linearly dependent.

4 0

3 years ago

Find the area of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximati

pochemuha

Answer: 8 (Pi - sqrt(3))

Discussion:

The area of the shaded region is that of the semicircle minus the area of the triangle..

Area of semicircle = 1/2 * Pi * R^2

Where R^2 is the square of the radius of the circle. In our case, R ( = OC)

= 4 so the semicircle area is

(1/2) * Pi * (4^2) = (1/2) * Pi * 16 = 8 Pi

Area of triangle.

First of all, angle ACB is a right angle ( i.e. 90 degrees).

* This is the Theorem of Thales from elementary Plane Geometry. *

so by Pythagoras

AC^2 + BC^2 = AB^2

But CB = 4 (given) and AB = 4*2 = 8 ( the diameter is twice the radius).

Substituting these in Pythagoras gives

AC^2 + 4^2 = 8^2 or

AC^2 = 8^2 - 4^2- = 64 - 16 = 48

Hence AC = sqrt(48) = sqrt (16*3) = 4 * sqrt(3)

We are almost done! The area of the triangle is given by

(1/2) b * h = (1/2) BC * AC = (1/2) 4 * (4 * sqrt(3)) = 8 sqrt(3)

We conclude the area area of the shaded part is

8 PI - 8 sqrt(3) = 8 (Pi - sqrt(3))

Note that sqrt(3) is approx 1.7 so (PI - sqrt(3)) is a positive number, as it better well be!

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3 years ago

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Usimov [2.4K]

Observing 150 is 3 times 50
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3 years ago