What is the value of ? 2/15 ÷ 5/12

Please help me !! Answer both correctly for a brainliest and also a thanks :)

neonofarm [45]

The firs t res is x=4
the second is D

6 0

3 years ago

46. Merchandising If 12 items are in a dozen, 12 dozen are in a gross, and 12 gross are in a great-gross, how many items are in

Andrews [41]

Answer: 1728

Step-by-step explanation: You multiply 12 times 12 because we have to find how many items are in a 12 dozen first which is 144 because there are 12 items in one dozen. We already know that there are 144 items in 12 dozens and there are 12 dozens in a gross, which is 144 items in total. Now we have to multiply 144 times 12 to find how many items are in a great gross. 144 times 12 is 1728.

3 0

4 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

3 years ago

Patrons (P) Revenue (R)

kari74 [83]

The linear model for the data is expressed as: R = 20p - 160.

<h3>How to Write a Linear Model?</h3>

Using two pairs of values from the table values, say, (32, 480) and (33, 500), find the unit rate (m).

Unit rate (m) = (500 - 480)/(33 - 32) = 20/1

Unit rate (m) = 20.

Substitute (p, R) = (32, 480) and m = 20 into R = mp + b to find b

480 = 20(32) + b

480 = 640 + b

480 - 640 = b

b = -160

To write the linear model, substitute m = 20 and b = -160 into R = mp + b:

R = 20p - 160

Learn more about linear model on:

brainly.com/question/4074386

#SPJ1

5 0

2 years ago

E is the midpoint of DF, DE = 2x + 4, and EF = 3x + 1. Find DE, EF, and DF.

Agata [3.3K]

When there is a midpoint in a line segment, the two segments that are then produced have to be equal. So that means that because E is a midpoint of DF, DE=EF.
Set up the equation.
DE=EF
2x+4=3x+1
Then solve for x.
x=3
Then plug it back in the equation to find DE and EF.
DE=2x+4=2(3)+4=10
Because DE=EF then EF is also 10.
DF is made from the addition of DE and EF so DF is twice EF.
DF= 20

DE= 10, EF= 10, DF= 20

Hope this helps.

8 0

3 years ago