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Paha777 [63]
3 years ago
9

Alex mows a lawn that measures 800 square feet in

Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0

First thing we gotta do is find the unit rate:

800 divided by 2.5 = 320

And now your problem is complete! The unit rate or square foot per hour is 320 square feet per hour!

Hope this helps, if not, comment below please!!!

You might be interested in
Can someone please help with question 38?
WARRIOR [948]
All you need to do is multiply all of the numbers together.
so 8x8x6x5x12
8x8x6x5x12=<span>23,040
so the surface area is </span><span>23,040</span>
3 0
3 years ago
1. In ABC, A=45 degrees, B=75 degrees, and BC=12. What is the length of AB?
Aleksandr-060686 [28]

We know that sum of the interior angles of any triangle equal 180° :

So :

A + B + C = 180°

45° + 75° + C = 180°

120° + C = 180°

Both sides minus ( 120°) :

C = 180° - 120°

C = 60°

_________________________________

According to the theorem of sinuses :

\frac{AB}{ \sin(C) } =  \frac{BC}{ \sin(A) } \\

\frac{AB}{ \sin(60) } =  \frac{12}{ \sin(45) } \\

\frac{AB}{ \frac{ \sqrt{3} }{2} } =  \frac{12}{ \frac{ \sqrt{2} }{2} } \\

Multiply \:  both \:  sides \:  by \:   \frac{ \sqrt{3} }{2}

AB =  \frac{24}{ \sqrt{2} } \times  \frac{ \sqrt{3} }{2} \\

AB =  \frac{12 \sqrt{3} }{ \sqrt{2} } \\

AB =  \frac{2 \times 6 \sqrt{3} }{ \sqrt{2} } \\

AB =  \frac{ ({ \sqrt{2} })^{2} \times 6 \sqrt{3}  }{ \sqrt{2} }  \\

AB =  \frac{ \sqrt{2} \times  \sqrt{2} \times 6 \sqrt{3}   }{ \sqrt{2} } \\

AB =  \frac{ \sqrt{2} \times  6 \sqrt{3}   }{1} \\

AB =  \sqrt{2} \times 6 \sqrt{3}

AB = 6 \sqrt{2 \times 3}

AB = 6 \sqrt{6}

_________________________________

I think this is the correct answer.

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

4 0
3 years ago
8 identical shirts cost a total $152. How much does 1 shirt cost?
sweet [91]
$19 because you divide 152 by 8
7 0
3 years ago
Read 2 more answers
4. Find the volume of the given solid bounded by the elliptic paraboloid z = 4 - x^2 - 4y^2, the cylinder x^2 + y^2 = 1 and the
Ilya [14]

Answer:

2.5π units^3

Step-by-step explanation:

Solution:-

- We will evaluate the solid formed by a function defined as an elliptical paraboloid as follows:-

                                  z = 4 - x^2 -4y^2

- To sketch the elliptical paraboloid we need to know the two things first is the intersection point on the z-axis and the orientation of the paraboloid ( upward / downward cup ).

- To determine the intersection point on the z-axis. We will substitute the following x = y = 0 into the given function. We get:

                                 z = 4 - 0 -4*0 = 4

- The intersection point of surface is z = 4. To determine the orientation of the paraboloid we see the linear term in the equation. The independent coordinates ( x^2 and y^2 ) are non-linear while ( z ) is linear. Hence, the paraboloid is directed along the z-axis.

- To determine the cup upward or downwards we will look at the signs of both non-linear terms ( x^2 and y^2 ). Both non-linear terms are accompanied by the negative sign ( - ). Hence, the surface is cup downwards. The sketch is shown in the attachment.

- The boundary conditions are expressed in the form of a cylinder and a plane expressed as:

                                x^2 + y^2 = 1\\\\z = 4

- To cylinder is basically an extension of the circle that lies in the ( x - y ) plane out to the missing coordinate direction. Hence, the circle ( x^2 + y^2 = 1 ) of radius = 1 unit is extended along the z - axis ( coordinate missing in the equation ).

- The cylinder bounds the paraboloid in the x-y plane and the plane z = 0 and the intersection coordinate z = 4 of the paraboloid bounds the required solid in the z-direction. ( See the complete sketch in the attachment )

- To determine the volume of solid defined by the elliptical paraboloid bounded by a cylinder and plane we will employ the use of tripple integrals.

- We will first integrate the solid in 3-dimension along the z-direction. With limits: ( z = 0 , z = 4 - x^2 -4y^2 ). Then we will integrate the projection of the solid on the x-y plane bounded by a circle ( cylinder ) along the y-direction. With limits: ( y = - \sqrt{1 - x^2} , y =  \sqrt{1 - x^2} ). Finally evaluate along the x-direction represented by a 1-dimensional line with end points ( -1 , 1 ).

- We set up our integral as follows:

                            V_s = \int\int\int {} \, dz.dy.dx

- Integrate with respect to ( dz ) with limits: ( z = 0 , z = 4 - x^2 -4y^2 ):

                           V_s = \int\int [ {4 - x^2 - 4y^2} ] \, dy.dx

- Integrate with respect to ( dy ) with limits: ( y = - \sqrt{1 - x^2} , y =  \sqrt{1 - x^2} )

                        V_s = \int [ {4y - x^2.y - \frac{4}{3} y^3} ] \, | .dx\\\\V_s = \int [ {8\sqrt{( 1 - x^2 )}  - 2x^2*\sqrt{( 1 - x^2 )} - \frac{8}{3} ( 1 - x^2 )^\frac{3}{2} } ] . dx

- Integrate with respect to ( dx ) with limits: ( -1 , 1 )

                       V_s =  [ 4. ( arcsin ( x ) + x\sqrt{1 - x^2} ) - \frac{arcsin ( x ) - 2x ( 1 -x^2 )^\frac{3}{2} + x\sqrt{1 - x^2}  }{2}  - \frac{ 3*arcsin ( x ) + 2x ( 1 -x^2 )^\frac{3}{2} + 3x\sqrt{1 - x^2}  }{3} ] | \limits^1_-_1\\\\V_s =  [ \frac{5}{2} *arcsin ( x ) + \frac{5}{3}*x ( 1 -x^2 )^\frac{3}{2} + \frac{5}{2} *x\sqrt{1 - x^2} ) ] | \limits^1_-_1\\\\V_s =  [ \frac{5\pi }{2}  + 0 + 0 ] \\\\V_s = \frac{5\pi }{2}

Answer: The volume of the solid bounded by the curves is ( 5π/2 ) units^3.

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8 0
3 years ago
P=2l+2w <br><br> Rearrange the formula above to isolate the variable w.
muminat
2w = P -2l
w = (P/2) -l

hope this helps :) (please mark brainliest if correct!)
4 0
3 years ago
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