7.25 cm plus 7.2 cm plus 8cm equals what

A submarine dives as shown in the diagram.

Marianna [84]

Answer:

Tan x = 125/500= 0.25

so x = arctan (0.25)

=14 degree

8 0

3 years ago

What is the slope of the line shown on the graph? A)3 B)1/3 C)−3 D)−1/3

MatroZZZ [7]

Answer:

1/3

Step-by-step explanation:

The slope of the line is found by

m = (y2-y1)/(x2-x1)

we have two point ( 0,0) and ( 3,1)

= ( 1-0)/( 3-0)

= 1/3

4 0

2 years ago

Trisha and Byron are washing and vacuuming cars to raise money for a class trip. Trisha raised $38 washington 5 cars and vacuumi

ANEK [815]

Answer:

They charged $6 for washing and $2 for vacuuming the car.

Step-by-step explanation:

Let the charge of washing a car be 'x'.

And also let the charge of vacuuming a car be 'y'.

Now according to question, Trisha raised $38 washing 5 cars and vacuuming 4 cars.

So framing the above sentence in equation form, we get;

$5x+4y=38\ \ \ \ \ equation\ 1$

Again according to question, Byron raises $28 by washing 4 cars and vacuuming 2 cars.

So framing the above sentence in equation form, we get;

$4x+2y=28\ \ \ \ \ equation\2$

Now multiplying equation 2 by '2', we get;

$2(4x+2y)=28\times2\\\\8x+4y=56\ \ \ \ equation\ 3$

Now we subtract equation 1 from equation 3 and get;

$(8x+4y)-(5x+4y)=56-38\\\\8x+4y-5x-4y=18\\\\3x=18\\\\x=\frac{18}{3}=6$

Now substituting the value of 'x' in equation 1, we get;

$5x+4y=38\\\\5\times6+4y=38\\\\30+4y=38\\\\4y=38-30=8\\\\y=\frac{8}{4}=2$

Hence They charged $6 for washing and $2 for vacuuming the car.

8 0

2 years ago

Twenty percent of the 640 pitches Kendall threw last year were strikes. Which table can be used to find the number of strikes sh

Ksju [112]

Answer:

1 out of every 5 throws were strikes.

Step-by-step explanation:

20% is one fifth of 100%.

5 0

2 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

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

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

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