All categories

3 years ago

What is the radius of a circle if the area is 78.5 cm^2

Mathematics

1 answer:

Verizon [17]3 years ago

3 0

78.5= (3.14)r^2
25=r^2
5=r
So, the radius of a circle with an area is 78.5 cm^2 is 5 cm^2

You might be interested in

Number 3.

Sergio039 [100]

Answer:

see explanation

Step-by-step explanation:

∠ PQX = 90° ( angle in a semicircle )

∠ PXQ = 42° ( alternate angle theorem )

Angle between tangent and diameter is right, thus

∠ XPQ = 90° - 42° = 48°

7 0

3 years ago

A triangle has a base of 4/5 of a yard and a height of 2/3 of a yard. What is the area of this triangle?

Elena-2011 [213]

Answer:

8/15

Step-by-step explanation:

4/5 times 2/3

4 0

3 years ago

The following steps can be used to compute

Vlada [557]

1. Commutative property of multiplication

IM SO SORRY IF IM WRONG

8 0

3 years ago

The difference of the squares of two positive consecutive even integers is 7676. find the integers. use the fact that, if x rep

myrzilka [38]

Equation is (x+2)^2 - x^2 = 7676
solve the equation by first opening the bracket of (x+2)^2 by A^2 + 2AB +C^2

you'll end up with 4x + 4 = 7672
thus x = 1918 and the next number is 1920.

5 0

3 years ago

If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

$a^2 - b^2 = (a-b) (a+b)$

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

5 0

2 years ago