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4 years ago

I need help with this pls be nice and tell me the answer :)

Mathematics

2 answers:

Alecsey [184]4 years ago

8 0

M=-10 the first guy was correct

KIM [24]4 years ago

7 0

Answer:

m=-10

Step-by-step explanation:

just to make this easier ill subtract 2 from each side to get

2/3m = 1/6m -5

now multiply each side by 6 to get rid of denominators (this just makes it easier)

now in the next steps we simplify

4m = 1m -30

3m= -30

m=-10

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What is the measure of each interior angle of a regular pentagon? Enter your answer in the box.

MAVERICK [17]

Hello there!

Pentagon = 5 sides.

For regular pentagon: 360/5 = 72. 180 - 72 = 108 degrees at each interior.

I hope this helps.

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3 years ago

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One angle in an obtuse isosceles triangle measures 40°. What are the measures of the other two angles?

iren2701 [21]

"In an isosceles triangle two angles are the same and the third is different. However, the angles in the triangle must add up to 180. If the 100 degree angle was one of the 2 identical angles, then the angles in the triangle would add up to more than 200. Because of this the 100 degree angle must be the different angle."

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4 years ago

PLEASE HELP I WILL GIVE 50 POINTS

vodka [1.7K]

32 + 72 = 104
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your answer would be 8

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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:

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

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

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

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

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

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

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

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

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

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

Assume that blood pressure readings are normally distributed with a mean of 123 and a standard deviation of 9.6. If 144 people a

Vanyuwa [196]

Answer:

The probability is 0.9938

Step-by-step explanation:

In this question, we are asked to calculate the probability that the mean blood pressure readings of a group of people is less than a certain reading.

To calculate this, we use the z score.

Mathematically;

z = (mean - value)/(standard deviation/√N)

From the question, we can identify that the mean is 125, the value is 123 , the standard deviation is 9.6 and N ( total population is 144)

Let’s plug these values;

z = (125-123)/(9.6/√144) = 2.5

Now we proceed to calculate the probability with a s score less than 2.5 using statistical tables

P(z<2.5) = 0.9938

8 0

3 years ago