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3 years ago

Median? 93; 17; 85; 42; 15; 19; 78

1 answer:

6 0

The median is the number in the middle when all of the numbers are placed in order (from lowest to highest).

15; 17; 19; 42; 78; 85; 93

The answer in this case would be 42.

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the length , breadth and height of a cuboid is 2x^2y cm,5y^2z cm and 4z^2x cm. The volume of the cuboid is ...............

kondor19780726 [428]

For this case we have that by definition, the volume of a cuboid is given by:

$V = L * A * h$

Where:

L: It's the long

A: It is the width

h: It is the height

According to the problem data we have:

$L = 2x ^ 2y \ cm\\A = 5y ^ 2z \ cm\\h = 4z ^ 2x \ cm$

So, the volume of the cuboid is:

$V = (2x ^ 2y) (5y ^ 2z) (4z ^ 2x)\\V = 10x ^ 2y ^ 3z (4z ^ 2x)\\V = 40x ^ 3y ^ 3z ^ 3$

Finally, the volume of the cuboid is: $V = 40x ^ 3y ^ 3z ^ 3$

Answer:

$V = 40x ^ 3y ^ 3z ^ 3$

7 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

2 years ago

The given diagram shows the parts of a right triangle with an altitude to the hypotenuse. Using the two given measures, find th

almond37 [142]

I only could find part of the triangle sorry, but u1 is 2 radical 2

8 0

2 years ago

What is the formula of LSD in a cuboid?

7nadin3 [17]

Answer: The volume of a cuboid is given by the formula V = LWH, and the surface area of a cuboid is given by the formula SA = 2lh + 2wh + 2lw where l = length, w = width, and h = height.

And i was just learning this a few days back lol

8 0

3 years ago

Choose the equation that represents the line that passes through the point (-1, 6) and has a slope of -3.

stich3 [128]

Answer:

Equation of line is $\mathbf{y=-3x+3}$