Write the standard equation of a circle with center (−4, 0) and radius 3.

The standard equation of a circle:

We have the center (-4, 0) → h = -4 and k = 0, and the radius r= 3. Substitute:

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.