What is the apparent solution to the system of equations graphed above?

What are the coordinates of the circumcenter of a triangle with vertices A(−3, 3) , B(−1, 3) , and C(−1, −1) ?

Shtirlitz [24]

Hello Missjay12,

Step 1 :
From the coordinates we have to find the slopes and midpoints of the lines.

Step 2 :
Find the slope of the bisectors and now find the equations of two lines by using slope and one mid points.

Step 3 :
Solve any two pair of equations and find the intersection point.

Step 4 :
The point thus found is the circumcenter of the given triangle.

so<span> the coordinates of the circumcenter of a triangle with vertices A(−3, 3) , B(−1, 3) , and C(−1, −1)</span> is, -4, 0.

8 0

3 years ago

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Explain how you could write a quadratic function in factored form that would have a vertex with an x-coordinate of 3 and two dis

KATRIN_1 [288]

The vertex is on the axis of symmetry, so the axis of symmetry is x = 3. Find any two x-intercepts that have the equivalent distance from the axis of symmetry. Use those x-intercepts to write factors of the function by subtracting their values from x. For example, 2 and 4 are each 1 unit from x = 3, so f(x) = (x – 2)(x – 4) is a possible function.

5 0

3 years ago

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A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

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-21(+5-8)7+-90:-6= yesssssssssssssssssssssssssssssssssssssssssssssssssss

hodyreva [135]

Answer:

yeeeeeeeees

Step-by-step explanation:

yeeeeeeeeeeees

8 0

2 years ago

Help! If u get it i will mark brainliest!

konstantin123 [22]

Answer:

Lead-the-market pay strategies. An employer may choose to establish an internal compensation strategy that is in excess of the pay rates in the prevailing marketplace. This compensation strategy may increase the supply of candidates, increase selection rates of qualified applicants, decrease employee turnover, increase morale and productivity, or prevent unionization efforts. However, prior to implementing a lead compensation strategy, an organization should carefully consider what benefits it expects to realize from such a strategy, keeping in mind that this type of structure has the greatest propensity of increasing overall labor costs.

Step-by-step explanation:

6 0

3 years ago

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