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3 years ago

To the nearest tenth, what is the area of a circle whose diameter is 9 feet?

Mathematics

2 answers:

diamong [38]3 years ago

8 0

A=πr squared.

9 divided by 2 = 4.5.

4.5 times 4.5 = 20.25.

A=(3.14) (20.25).

A=63.6.

FromTheMoon [43]3 years ago

5 0

63.6 is the answer I got.

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Solve for W <br> 4w^2+9w= -2

7nadin3 [17]

Answer:

$w_1=-\frac{1}{4},w_2=-2}$

Step-by-step explanation:

$4w^2+9w=-2$

$4w^2+9w+2=0$

$w^2+9w+8=0$

$(w+1)(w+8)=0$

$(4w+1)(w+2)=0$

$w_1=-\frac{1}{4},w_2=-2}$

3 0

2 years ago

20 points and will give brainliest!!

FromTheMoon [43]

Answer:

$C=0.65M+22.65$

Step-by-step explanation:

The standard charge includes an initial fee and plus an addditional fee for each mile driven and is given by:

$S=0.40M+17.75$

The insurance charge is given by:

$I=0.25M+4.90$

So, the total charge C will be the sum of the standard charge S and the insurance charge I:

$C=S+I$

Substitute:

$C=(0.40M+17.75)+(0.25M+4.90)$

Rearrange:

$C=(0.40M+0.25M)+(17.75+4.90)$

Add:

$C=0.65M+22.65$

And we have acauired our equation relating C to M!

4 0

3 years ago

Please help I need this please help please just me help

Nikitich [7]

Answer:

It would take 4 minutes.

Step-by-step explanation:

Step 1: divide 63 by 3 to get how many hotdogs the champion can eat in a minute which equals 21

Step 2: Divide 84 by 21 to get 4 minutes

3 0

2 years ago

6) the circular base of a hemisphere of radius 2 rests on the base of a square pyramid of height 6. the hemisphere is tangent to

Rasek [7]

The length of the square base is thus $2 x 3\sqrt{2}/2 = 3\sqrt{2}$ = A

<h3>What is hemisphere?</h3>

Consequently, a hemisphere is a 3D geometric object that is made up of half of a sphere, with one side being flat and the other being a bowl-like shape. It is created by precisely cutting a spherical along its diameter, leaving behind two identical hemispheres.

EXPLANATION; Let ABCDE be the pyramid with ABCD as the square base. Let O and M be the center of square ABCD and the midpoint of side AB respectively. Lastly, let the hemisphere be tangent to the triangular face ABE at P.

Notice that triangle EOM has a right angle at O. Since the hemisphere is tangent to the triangular face ABE at P, angle EPO is also 90 degree. Hence, triangle EOM is similar to triangle EPO.

OM/2 = 6/EP

OM = 6/EP x 2

OM = $6\sqrt{6^2 - 2^2} x 2 = 3\sqrt{2}/2}$

The length of the square base is thus $2 x 3\sqrt{2}/2 = 3\sqrt{2}$ = A

To know more about Hemisphere, visit;

brainly.com/question/13625065?referrer=searchResults

#SPJ4

6 0

1 year ago

Polygon ABCD is dilated, rotated, and translated to form polygon A′B′C′D′. The endpoints of AB are at (0, -7) and (8, 8), and th

Anton [14]

Rotation and translation are rigid transformations, they don't change figure sizes. Dilation change figure sizes increasing or decreasing them by scale factor.

First, find AB and A'B' by the formula:

$AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}= \sqrt{(8-0)^2+(8-(-7))^2}=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17,\\ \\A'B'=\sqrt{(x_{B'}-x_{A'})^2+(y_{B'}-y_{A'})^2}= \sqrt{(2-6)^2+(1.5-(-6))^2}=\sqrt{4^2+7.5^2}=\sqrt{16+56.25}=\sqrt{72.25}=8.5.$

As you can see AB=2A'B'. This means that the segment AB was decreased twice to form segment A'B'. Then the scale factor is 1/2.

8 0

3 years ago