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3 years ago

The length of a rectangle measures 1 1/2 feet. The width measures 30 inches. Find the area of the rectangle.

Mathematics

1 answer:

aleksley [76]3 years ago

5 0

Answer:

540 in^2, or, equivalently, 3.75 ft^2

Step-by-step explanation:

Here, Area = A = (length)(width).

Converting 1 1/2 feet to inches, we get 18 inches.

Then A = (18 in)(30 in) = 540 in^2, or 3.75 ft^2

(Recall that 1 ft^2 = 144 in^2)

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3 years ago

Circle O and triangle OQR are shown below. What is the length of OQ?

zysi [14]

Answer:

Step-by-step explanation:

A circle is inscribed in an equilateral triangle PQR with centre O. If angle OQR = 30°, what is the perimeter of the triangle?

This is a circle inscribed in an equilateral triangle with side s.

If you are asking for the perimeter of PQR, it is 3s.

If you are asking for the perimeter of OQR, it is (3+23–√3)s

Since OR and SR are the hypotenuses of right triangles with adjacent side equal to ½ s, their length is ½s / cos 30° = (√3) /3.

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2 years ago

If m 1 = 140 and m 3 = 50, what is m 5

dolphi86 [110]

The answer is: m5 = -40.

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4 years ago

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3 years ago

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

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3 years ago

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