Evaluate M/q + np2 for m = 24, n = 6, p = 9, and q = 8.

The diameter of the larger semicircle is subtended by the two smaller semicircles, the small semicircle closer to the left of the internal circumference of the larger semicircle is shaded one while the one on the right is without color,

Therefore,

Diameter of larger semicircle = 2 × 10 = 20 cm

Based on the diagram, the shaded area is observed to be;

Shaded area = Area of semicircle formed by larger diameter or 20 cm + Area area formed by the small semicircle close to the right of the internal circumference of the larger semicircle - Area area formed by the other smaller semicircle

Since the diameter and therefore the areas of the two small semicircles are equal, we have;

Shaded area = Area formed by the complete larger semicircle = Area of semicircle formed by larger diameter or 20 cm

∴ Shaded area = π·D²/4 = π×20²/4 = 100×3.142 cm² = 314.2 cm².

6 0

3 years ago

In place value the Relationship between 2,000 and 20 is how many times

Katen [24]

Answer:

100

Step-by-step explanation:

20 times 100 equals 2000

6 0

3 years ago

A family wants to build a rectangular garden on one side of a barn. If 600 feet of fencing is available to use, then what is the

vovangra [49]

Answer: (A) $A=300l-l^{2}$

(B) Length varies between 1 and 150

(C) Largest area is 22500ft²

Step-by-step explanation: Suppose length is l and width is w.

The rectangular garden has perimeter of 600ft, which is mathematically represented as

$2l+2w=600$

Area of a rectangle is calculated as

$A=lw$

Now, we have a system of equations:

$2l+2w=600$

$A=lw$

Isolate w, so we have l:

$2w=600-2l$

w = 300 - l

Substitute in the area equation:

A = l(300 - l)

A = 300l - l²

(A) Function of area in terms of length is given by A = 300l - l²

(B) The practical domain for this function is values between 1 and 150.

(C) For the largest area, we need to determine the largest garden possible. For that, we take first derivative of the function:

A' = 300 - 2l

Find the values of l when A'=0:

300 - 2l = 0

2l = 300

l = 150

Replace l in the equation:

w = 300 - 150

w = 150

Now, calculate the largest area:

A = 150*150

A = 22500

The largest area the fence can enclose is 22500ft².

4 0

3 years ago