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2 years ago

A small explanation would be really appreciated

Mathematics

1 answer:

shutvik [7]2 years ago

7 0

Step-by-step explanation:

the 2 angles marked with orange and green are equal becos the question said that the indicated angles are congruent meaning same deg. so 180 - that given angle = the 2 marked angles (based on the rule known as angles on a str line are supplementary)

since the 2 marked angles are same, you can say that d and k are parallel due to the rule of alternate angles

Topic: angle properties

If you like to venture further, feel free to check out my insta (learntionary). It would be best if you could give it a follow. I'll be constantly posting math tips and notes! Thanks!

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Mary throws her Grandma’s favorite crystal dish out the second floor window, which is 6 meters from the ground. She is really up

wlad13 [49]

we know that

acceleration due to gravity is 9.8m/s^2

so, we get

$a(t)=9.8$

and acceleration will always be constant

we know that

integral of acceleration is velocity

so, we can integrate both sides

$\int \:a\left(t\right)dt=\int \:9.8dt$

$v(t)=9.8t+C$

we are given

v(0)=35m/s

we can use it and find C

$v(0)=9.8*0+C$

$35=9.8*0+C$

$C=35$

now, we can plug it back

$v(t)=9.8t+35$

we know that

integral of velocity is height

we can integrate it again

$\int \:v\left(t\right)dt=\int \:\left(9.8t+35\right)dt$

$s(t)=\int \:\left(9.8t+35\right)dt$

$s(t)=4.9t^2+35t+C$

now, we have

s(0)=6

we can use it and find C

$s(0)=4.9(0)^2+35(0)+C$

$6=4.9(0)^2+35(0)+C$

now, we can plug back C

$C=6$

and we get

$s(t)=4.9t^2+35t+6$ .............Answer

4 0

3 years ago

Read 2 more answers

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

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