What is the rise of a line that has a run of 20 and a slope of 1

Find the area enclosed by the figure. Use 3.14 for π. (The figure is not to scale).<br><br>

gtnhenbr [62]

Answer:

97.27 yd²

Step-by-step explanation:

First let's find the area of the right triangle on the left side. To do that, multiply 6x5, then divide the product by 2, since a right triangle is half a rectangle.

6x5=30

30/2=15

Area of triangle=15 yd²

Find area of the middle rectangle. So multiply the length by the width.

9x6=54

Area of rectangle=54 yd²

Find area of the half circle. Formula for area of circle is $A=\pi r^{2}$ . Pi is about 3.14, and radius ( $r$ ) is 3. Radius is half of the diameter of circle, and the 6 on the left says the diameter of circle.

$\pi$ x $3^{2}$ ≈28.27

Area of circle is about 28.27 yd²

Add up the three areas.

15+54+28.27=97.27

3 0

3 years ago

Read 2 more answers

An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh

natima [27]

Answer:

Speed of the plane in still air: $779\; {\rm km \cdot h^{-1}}$ .

Windspeed: $100\; {\rm km \cdot h^{-1}}$ .

Step-by-step explanation:

Assume that $x\; {\rm km \cdot h^{-1}}$ is the speed of the plane in still air, and that $y\; {\rm km \cdot h^{-1}}$ is the speed of the wind.

When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be $(x - y)\; {\rm km \cdot h^{-1}}$ .
When this plane is travelling in the same direction as the wind, the ground speed of this plane would be $(x + y)\; {\rm km \cdot h^{-1}}$ .

The question states that when going against the wind ( $v = (x - y)\; {\rm km \cdot h^{-1}}$ ,) the plane travels $6111\; {\rm km}$ in $9\; {\rm h}$ . Hence, $9\, (x - y) = 6111$ .

Similarly, since the plane travels $7911\; {\rm km}$ in $9\; {\rm h}$ when travelling in the same direction as the wind ( $v = (x + y)\; {\rm km \cdot h^{-1}}$ ,) $9\, (x + y) = 7911$ .

Add the two equations to eliminate $y$ . Subtract the second equation from the first to eliminate $x$ . Solve this system of equations for $x$ and $y$ : $x = 779$ and $y = 100$ .

Hence, the speed of this plane in still air would be $779\; {\rm km \cdot h^{-1}}$ , whereas the speed of the wind would be $100\; {\rm km \cdot h^{-1}}$ .

3 0

2 years ago

Can you tell me which one plz ?

Andreas93 [3]

Just saying you have a 25% chance of getting it right if you guess

7 0

3 years ago

Need to pass whats 1+3-4+24-23=

SpyIntel [72]

Answer:

the answer is 5

Step-by-step explanation:

4 0

3 years ago

What is the fourth term in the expansion of (x + 2y)^8?

Hatshy [7]

Base on my calculation and the simplification of the form and finding its 4th power of the x+2y, i came up with a solution and finally found an answer of B. 448x^5y^3, I hope i answer your questions directly and please feel free to ask for more and ask for clarifications

4 0

3 years ago