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2 years ago

Simplify square root

Mathematics

1 answer:

NemiM [27]2 years ago

7 0

Answer:

Step-by-step explanation:

look up rules of multiplying radicals.

You can multiply the radicands (the number inside the square root) then take the square root of the new number.

3*21 = 63

Factor 63 into 3*3*7

Pull out 3^2 from inside the square root

leaves you with 3sqrt7 or answer C

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The picture shows a sandwich box:

loris [4]

Using the Pythagorean theorem we can find the height of the triangle:

height = √(14^2 - 12^2)

height = √(196 - 144)

height = √52

height = 7.2

The answer would be 7.2 centimeters.

7 0

2 years ago

Read 2 more answers

Find the volume of the solid that lies between planes perpendicular to the x-axis at x = pi/4 and x = 5pi/4. The cross-sections

svp [43]

Answer:

The volume of the solid = π²

Step-by-step explanation:

As per the given data of the questions,

The diameter of each disk is

D = 2 sin(x) - 2 cos(x)

So its radius is

R = sin(x) - cos(x).

The area of each disk is

$= \pi \times(sinx-cosx)^{2}$

$= \pi \times [sin^{2}(x) - 2 sin(x) cos(x) + cos^{2}(x)]$

$= \pi[1-2sin(x)cos(x)]$

$= \pi[1-sin(2x)]$

Now,

Integrate from $x=\frac{\pi}{4}\ \ \ to\ \ \ x=\frac{5\pi}{4}$ , we get volume:

$V=\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \pi[1-sin(2x)]dx$

After integrate without limit we get

$V=\pi[x+\frac{cos2x}{2}]$

Now after putting the limit, we get

V = π²

Hence, the required volume of the solid = π²

3 0

3 years ago

Select whether the equation has a solution or not.<br> √x = 7

Ierofanga [76]

Answer:

x = 49

The equation does have a solution

Step-by-step explanation:

The sqrt of x = 7, does have a solution. Simply remove the radical, and raise each side by the index of the radical.

In more basic terms, what number has the sqrt of 7? 7^2 is equal to 49!

6 0

3 years ago

Select the correct answer.

Brums [2.3K]

ответ: д
-1/2 * 16 = -8
или 16/(-2) = -8

4 0

1 year ago

Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

$\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA$

7 0

2 years ago