A golfer has a 4 day tournament what is the final score

You measure a piece of rope and find its length to be 12m . Find the percent error of this measurement

MrRissso [65]

percentage error is .5/12 = 4.17%

Hope this helps!

~LENA~

6 0

2 years ago

What polygon is formed when 2 equilateral triangles that are congruent and share a side

zalisa [80]

You form a rhombus.

Draw an equilateral triangle, and then mirror it around its base. All the sides are clearly equal, because they are all sides of the same equilateral triangle.

Moreover, the top and bottom angle are 60°, because they are angles of an equilateral triangle, while the left and right angles are 120, because they're 60+60 angles.

7 0

2 years ago

) Which point on the number line shows the position of squared 77

a_sh-v [17]

The root of 77 is 8.774 so the answer is C

Answer: C

8 0

1 year ago

A storage tank will have a circular base of radius r and a height of r. The tank can be either cylindrical or hemispherical (ha

Neko [114]

Answer:

Volume: $\frac{2}{3}\pi r^3$

Ratio: $\frac{2}{9}r$

Step-by-step explanation:

First of all, we need to find the volume of the hemispherical tank.

The volume of a sphere is given by:

$V=\frac{4}{3}\pi r^3$

where

r is the radius of the sphere

V is the volume

Here, we have a hemispherical tank: a hemisphere is exactly a sphere cut in a half, so its volume is half that of the sphere:

$V'=\frac{V}{2}=\frac{\frac{4}{3}\pi r^3}{2}=\frac{2}{3}\pi r^3$

Now we want to find the ratio between the volume of the hemisphere and its surface area.

The surface area of a sphere is

$A=4 \pi r^2$

For a hemisphere, the area of the curved part of the surface is therefore half of this value, so $2\pi r^2$ . Moreover, we have to add the surface of the base, which is $\pi r^2$ . So the total surface area of the hemispherical tank is

$A'=2\pi r^2 + \pi r^2 = 3 \pi r^2$

Therefore, the ratio betwen the volume and the surface area of the hemisphere is

$\frac{V'}{A'}=\frac{\frac{2}{3}\pi r^3}{3\pi r^2}=\frac{2}{9}r$

6 0

2 years ago

20 PTSS !!!<br> PLEASE HELPP !!

butalik [34]

Because it is extremely hard to find the area of this figure all together, it would be in our best interest to split this figure up into three different pieces: the two horizontal rectangles, and the verticle rectangle. We can find the area of all three and add them up. Be aware that there are two different ways that you can break this figure up, As shown in the attachments. I will be using the first image (the one with the tall horizontal rectangles, NOT the almost-squares).

So, we see that we have enough information to solve for the area of the left-most rectangle. Area = lw. 10 x 4 = 40, so the area is 40. Next, we have to notice, that the horizontal rectangles are also the same, so both of the areas of the two horizontal rectangles are 40.

Now, we can find the middle rectangle. We know that the length of the entire thing is 18, but it is taken up by 8 (4+4) of the horizontal triangles, so 18-8=10, so the length Is 10. We also know that the height of the horizontal rectangles is 10, so 10-3=7. Our dimensions for the rectangle are 10x7 or 70 square units. If we add them all together, 40+40+70=150.

The area is 150 square units

8 0

2 years ago