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3 years ago

10

WILL GIVE BRAINLIEST! Six numbers form an arithmetic sequence with a common difference of 4. The sum of these numbers equals to

12. What are the six numbers?

1 answer:

GrogVix [38]3 years ago

3 0

Answer: -8,-4,0,4,8,12

Step-by-step explanation:

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Write the equation of the given circle.<br><br> center (1, -5)<br> radius of 10

Tamiku [17]

Answer:

314

Step-by-step explanation:

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3 years ago

Please help!! will mark branliest

Mashcka [7]

Answer:

a. After the first bounce, the ball will be at 85% of 8 ft. After 2 bounces, it'll be at 85% of 85% of 8 feet. After 3 bounces, it'll be at (85% of) (85% of) (85% of 8 feet). You can see where this is going. After n bounces the ball will be at

$h(n) = (85\%)^n 6ft = (\frac{17}{20})^n(6ft\times 12\frac{in}{ft}) = (\frac{17}{20})^n\times72'$

b. After 8 bounces we can apply the previous formula with n = 8 to get

$h(8)=(\frac{17}{20})^8\times 72' = 19.62'$

c. The solution to this point requires using exponential and logarithm equations; a more basic way would be trial and error using the previous $h(n)$ increasing the value of n until we find a good value. I recommend using a spreadsheet for that; the condition will lead to the following inequality:

$1>(\frac{17}{20})^n\times 72$ Let's first isolate the fraction by dividing by 72.

$(\frac{17}{20})^n$ Now, to get numbers we can plug in a calculator, let's take the natural logarithm of both sides:

$ln ((\frac{17}{20})^n) < ln \frac 1{72} \rightarrow n\times ln (\frac{17}{20}) < -ln72$ . Now the two quantities are known - or easy to get with any calculator, replacing them and solving for n we get:

$-(0.16)n < - 4.27 \rightarrow n>26.31$ Now, since n is an integer - you can't have a fraction of a bounce after all, you pick the integer right after that, or n>27.

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3 years ago

I need help!! ASAP!!

ohaa [14]

Answer:your answer is c

Step-by-step explanation:if you want to be friends we can just friend me

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3 years ago

The list shows numbers in order from least to greatest.

Andrei [34K]

Answer:

999

Step-by-step explanation:

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3 years ago

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