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FinnZ [79.3K]
3 years ago
12

Error Analysis! Answer both parts of the question.

Mathematics
1 answer:
Maurinko [17]3 years ago
8 0

Answer:

Kaitlin's answer is wrong because she is supposed to by dividing. When trying to get rid of a number from a variable you have to divide the number by itself and whatever you do to one side you have to do to the other side

so you would do 5/5 - cancels out

15/5=3

y=3

Step-by-step explanation:

have a great day :D

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BabaBlast [244]

Answer:

8

Step-by-step explanation:

(16^3/2)^1/2

We know that a^ b^c = a^(b*c)

16 ^(3/2*1/2)

16 ^3/4

Now rewriting 16 as 2^4

2^4^3/4

2 ^ (4*3/4)

2^3

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6 0
4 years ago
A line passes through (2, −1) and (4, 5).
liberstina [14]
The answer to the first question is B
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How can you determine the period of a graph
Vlad1618 [11]

Answer:

You need to find the cycle. It will be easier to show if a graph with it was shown

Step-by-step explanation:

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3 years ago
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Find the area of the surface. The part of the sphere x2 + y2 + z2 = a2 that lies within the cylinder x2 + y2 = ax and above the
sineoko [7]

Answer:

The area of the sphere in the cylinder and which locate above the xy plane is \mathbf{ a^2 ( \pi -2)}

Step-by-step explanation:

The surface area of the sphere is:

\int \int \limits _ D \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )   } \ dA

and the cylinder x^2 + y^2 =ax can be written as:

r^2 = arcos \theta

r = a cos \theta

where;

D = domain of integration which spans between \{(r, \theta)| - \dfrac{\pi}{2} \leq \theta  \leq \dfrac{\pi}{2}, 0 \leq r \leq acos \theta\}

and;

the part of the sphere:

x^2 + y^2 + z^2 = a^2

making z the subject of the formula, then :

z = \sqrt{a^2 - (x^2 +y^2)}

Thus,

\dfrac{\partial z}{\partial x} = \dfrac{-2x}{2 \sqrt{a^2 - (x^2+y^2)}}

\dfrac{\partial z}{\partial x} = \dfrac{-x}{ \sqrt{a^2 - (x^2+y^2)}}

Similarly;

\dfrac{\partial z}{\partial y} = \dfrac{-2y}{2 \sqrt{a^2 - (x^2+y^2)}}

\dfrac{\partial z}{\partial y} = \dfrac{-y}{ \sqrt{a^2 - (x^2+y^2)}}

So;

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = \sqrt{\begin {pmatrix} \dfrac{-x}{\sqrt{a^2 -(x^2+y^2)}} \end {pmatrix}^2 + \begin {pmatrix} \dfrac{-y}{\sqrt{a^2 - (x^2+y^2)}}   \end {pmatrix}^2+1}\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = \sqrt{\dfrac{x^2+y^2}{a^2 -(x^2+y^2)}+1}

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = \sqrt{\dfrac{x^2+y^2+a^2 -(x^2+y^2)}{a^2 -(x^2+y^2)}}

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = \sqrt{\dfrac{a^2}{a^2 -(x^2+y^2)}}

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = {\dfrac{a}{\sqrt{a^2 -(x^2+y^2)}}

From cylindrical coordinates; we have:

\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )}  = {\dfrac{a}{\sqrt{a^2 -r^2}}

dA = rdrdθ

By applying the symmetry in the x-axis, the area of the surface will be:

A = \int \int _D \sqrt{ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2+1} \ dA

A = \int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} \int ^{a cos \theta}_{0} \dfrac{a}{\sqrt{a^2 -r^2 }} \ rdrd \theta

A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 -r^2} \end {bmatrix}^{a cos \theta}_0 \ d \theta

A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 - a^2cos^2 \theta} + a \sqrt{a^2 -0}} \end {bmatrix} d \thetaA = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \ sin \theta +a^2 } \end {bmatrix} d \theta

A = 2a^2 [ cos \theta + \theta ]^{\dfrac{\pi}{2} }_{0}

A = 2a^2 [ cos \dfrac{\pi}{2}+ \dfrac{\pi}{2} - cos (0)- (0)]

A = 2a^2 [0 + \dfrac{\pi}{2}-1+0]

A = a^2 \pi - 2a^2

\mathbf{A = a^2 ( \pi -2)}

Therefore, the area of the sphere in the cylinder and which locate above the xy plane is \mathbf{ a^2 ( \pi -2)}

6 0
3 years ago
Burros are amphibians, true or false?
Natasha_Volkova [10]

The given statement is False.

Amphibians are the animals that can live both in water and on the water. For example frog.

Burros are a pack of small donkey. They are found in North American deserts.

They don't live in water. Hence, Burros are not amphibians.

Therefore, the given statement is False.

3 0
4 years ago
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