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Dafna1 [17]
3 years ago
15

What is the variable 4t+8-5c

Mathematics
1 answer:
Svetlanka [38]3 years ago
7 0

Answer: The variable is t and c.

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Find the coordinates of the points that are 10 units away from the origin and have a y- coordinate equal to -8
MrRissso [65]

Answer:

(-6, -8) and (6, -8)

Step-by-step explanation:

Think of this as a right triangle

The hypotenuse is 10

The "y" leg is |-8| = 8

x² + 8² = 10²

x² + 64 = 100

x² = 36

x = ±6

two points

(-6, -8) and (6, -8)

8 0
3 years ago
Does 8 has more factors than 9
gayaneshka [121]
*have 
and yes 
8 factor
   1,2,4,8
9 factor
    1,3,9
5 0
3 years ago
Read 2 more answers
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
Can you help me fill in the missing blanks thank you!!
Natasha_Volkova [10]
You just have to find what number times 9 equals your original equation so 9 and -2 are -18 and 9 and 3 are 27 so your distributed equation would be:

9(-2x+3)
5 0
3 years ago
X=15-y<br> y=13-z<br> z=12-x
Ksju [112]

Answer:

x = 7; y = 8; z = 5

Step-by-step explanation:

x = 15 - y

y = 13 - z

z = 12 - x

The first equation is solved for x. Substitute x in the third equation with 15 - y from the first equation.

z = 12 - (15 - y)

z = 12 - 15 + y

z = y - 3       (Equation 1)

Now substitute y - 3 for z in the original second equation.

y = 13 - z

y = 13 - (y - 3)

y = 13 - y + 3

2y = 16

y = 8

Substitute 8 for y in Equation 1.

z = y - 3

z = 8 - 3

z = 5

Substitute 5 for z in the original third equation.

5 = 12 - x

-7 = -x

x = 7

Solution: x = 7; y = 8; z = 5

8 0
3 years ago
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