_12X+18(-11) solve it

The numerator of a given fraction is 4 less than the denominator. if 3 is subtracted from the numerator and 5 is added to the de

Sergeu [11.5K]

Answer:

Step-by-step explanation:Given that the numerator of a given fraction is 4 less than its denominator.

Also given that 3 is subtracted from the numerator and 5 is added to the denominator, the fraction becomes one by fourth .

Let the fraction be

Since the numerator of a given fraction is 4 less than its denominator we have,

Numerator=Denominator-4

⇒ a=b-4

Since 3 is subtracted from the numerator and 5 is added to the denominator, the fraction becomes one by fourth we have

4(a-3)=1(b+5)

4a-12=b+5

4a-b=17

4(b-4)-b=17 ( ∵ a=b-4)

4b-16-b=17

3b=17+16

3b=33

⇒ b=11

Now put b=11 in a=b-4 we get

a=11-4

⇒ the fraction is a/b=7/11

3 0

2 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

Q and R are not mutually exclusive events. p (q) = 3/5 p(r) = 1/3 and p ( q and r)= 1/5 find P(Q or R).

V125BC [204]

Recall the inclusion/exclusion principle:

$\mathbb P(Q\cup R)=\mathbb P(Q)+\mathbb P(R)-\mathbb P(Q\cap R)$

which means

$\mathbb P(Q\text{ or }R)=\dfrac35+\dfrac13-\dfrac15=\dfrac{11}{15}$

5 0

2 years ago

The sum of eight times a number and three is fifty-nine. Find the number.

jarptica [38.1K]

Answer:

x=7

Step-by-step explanation:

8x +3 = 59

8x= 56

/8

x = 7

6 0

2 years ago

Bella’s car payment is $321.00 a month for a five year period. What will she have paid for her car when she has paid her finals

Naya [18.7K]

Answer:

$19,260

Step-by-step explanation:

1 year = 12 months

5 × 12 = 60

5 years = 60 months

$321 × 60 = $19,260

8 0

1 year ago