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3 years ago

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:

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

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

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

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

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

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

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

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

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

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