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tatuchka [14]
3 years ago
7

Mathematics, Which choice, and explain plz

Mathematics
2 answers:
Tcecarenko [31]3 years ago
5 0

Answer:

F. 60 mph

Step-by-step explanation

Use this equation

(180*2)/(2+4) and then you will get 60.

Alex787 [66]3 years ago
5 0

Answer:

F. 60 mph

Step-by-step explanation:

Use this equation

(180*2)/(2+4) =

= 360/6 =

= 60

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How can the solution be located?

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Please help me i need to find the area of shaded region please help me​
Montano1993 [528]

Answer:

Part 1) A=60\ ft^2

Part 2) A=80\ cm^2

Part 3) A=96\ m^2

Part 4) A=144\ cm^2

Part 5) A=13\ m^2

Part 6) A=(49\pi -33)\ in^2

Step-by-step explanation:

Part 1) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior rectangle

The area of rectangle is equal to

A=bh

where

b is the base of rectangle

h is the height of rectangle

so

A=(12)(7)-(8)(3)\\A=84-24\\A=60\ ft^2

Part 2) we know that

The shaded region is equal to the area of the complete rectangle minus the area of the interior square

The area of square is equal to

A=b^2

where

b is the length side of the square

so

A=(12)(8)-(4^2)\\A=96-16\\A=80\ cm^2

Part 3) we know that

The area of the shaded region is equal to the area of four rectangles plus the area of one square

so

A=4(4)(5)+(4^2)\\A=80+16\\A=96\ m^2

Part 4) we know that

The shaded region is equal to the area of the complete square minus the area of the interior square

so

A=(15^2)-(9^2)\\A=225-81\\A=144\ cm^2

Part 5) we know that

The area of the shaded region is equal to the area of triangle minus the area of rectangle

The area of triangle is equal to

A=\frac{1}{2}(b)(h)

where

b is the base of triangle

h is the height of triangle

so

A=\frac{1}{2}(6)(7)-(4)(2)\\A=21-8\\A=13\ m^2

Part 6) we know that

The area of the shaded region is equal to the area of the circle minus the area of rectangle

The area of the circle is equal to

A=\pi r^{2}

where

r is the radius of the circle

so

A=\pi (7^2)-(3)(11)\\A=(49\pi -33)\ in^2

6 0
3 years ago
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