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3 years ago

Solve this equation using an algebraic method.

Mathematics

1 answer:

kupik [55]3 years ago

7 0

(x+4) (x-4) = 9
(x^2 - 16) = 9
x^2 = 25
x=5 & x = -5

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Gnoma [55]

Answer:

area of kite=d1d2/2

~3tu(4tu+10u)/2

12t^2u^2+30tu^2/2

=====area=6t^2u^2+15ut^2

c option

3 0

3 years ago

Please help answer the following question

Elena L [17]

Answer:

Step-by-step explanation:

The unmarked angle of the triangle is also 40o. The triangle is marked with two equal sides (isosceles).

<2 + 40 + 40 = 180 Combine the left

<2 + 80 = 180 Subtract 80 from both sides

<2 = 180 - 80

<2 = 100 degrees.

That's not the answer. We have to solve for x

<2 = x + 106

100 = x + 106 Subtract 106 from both sides

100 - 106 = x

x = - 6

8 0

3 years ago

Serhud [2]

A circle is characterized by radius, arc, sectors and circumference

The length of the major arc is $18.75\pi$
The radius of the circle is 15
The area of the shaded sector is $140.625 * \pi$

<h3>Length of the major arc</h3>

The given parameters are:

$C = 30\pi$ --- the circumference

$\theta = 225^o$ -- the center angle

The length of the major arc is calculated using:

$L = \frac{\theta}{360} * C$

So, we have:

$L = \frac{225}{360} *30\pi$

Evaluate

$L = 18.75\pi$

Hence, the length of the major arc is $18.75\pi$

<h3>The radius of the circle</h3>

The circumference is given as:

$C = 30\pi$

So, we have:

$2\pi r = 30\pi$

Divide through by 2pi

$r = 15$

Hence, the radius of the circle is 15

<h3>The area of the shaded sector</h3>

The area of a sector is:

$A = \frac{\theta}{360} * \pi r^2$

So, we have:

$A = \frac{225}{360} * \pi * 15^2$

Evaluate

$A = 140.625 * \pi$

Hence, the area of the shaded sector is $140.625 * \pi$