Assume ∠a and ∠b are acute
Answer:
when opening the bracket remember that the value outside the bracket will be used to multiply the value inside the bracket
Let's assume
length of box is l
width of box is w
we know that
area of rectangle = length*width
...................(1)
now, we are given
fencing length is 20 foot
fencing length will be equal to perimeter
and we know that

so, we get

now, we can solve for l

now, we can plug that in first equation

now, we can solve for w




now, we can find length

or

so, option-B..................Answer
6. Multiply the given equation by 20
... 20×1 in. = 20×2.54 cm
... 20 in = 50.8 cm
7. Multiply the equation (1 foot = 12 inches) by 25.
... 25×1 ft = 25×12 in
... 25 ft = 300 in
Now, multiply the equation for conversion to cm by 300
... 300×1 in = 300×2.54 cm
... 300 in = 762 cm
Using the transitive property of equality,
... 25 ft = 300 in = 762 cm
... 25 ft = 762 cm
_____
A "unit multiplier" is no doubt explained in your lesson. We assume it is a fraction whose value is 1, but that has different units in numerator and denominator. The equation 1 in = 2.54 cm gives rise to two unit multipliers, depending on which conversion you need or want:

Since you're converting <em>to</em> centimeters, you want the one with <em>cm</em> in the numerator.


We are given a circle with a partially shaded region. First, we need to determine the area of the whole circle. To do this, we need the measurement of the radius of the circle:
Use the Pythagorean theorem to solve for the other leg of the right triangle inside the circle:
5^2 = 3^2 + x^2
x = 4
The radius is 4 + 1 cm = 5 cm
So the area of the circle is A = pi*r^2
A = 3.14 * (5)^2
A = 25pi cm^2
To solve for the area of the shaded region:
Ashaded = Acircle - Atriangles
we need to solve for the area of the triangles:
A = 1/2 *b*h
A = 1/2 *6 * 5
A = 15 cm^2
Atriangles = 2 * 15
Atriangles = 30 cm^2
Ashaded = 25pi - 30
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