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jek_recluse [69]
3 years ago
5

Anybody have the answer to this?

Mathematics
1 answer:
bixtya [17]3 years ago
6 0
I assume that the segment that measures 13.5 cm is a tangent to the circle at the point we see. The side that measures x is a radius of the circle, so the side of the triangle that measures x and the side that measures 13.5 cm are perpendicular. The triangle is a right triangle.
One leg measures x.
One leg measures 13.5 cm.
The hypotenuse measures x + 8.45 cm.
With the lengths of the the 3 sides, we can use the Pythagorean theorem and solve for x.

x^2 + (13.5)^2 = (x + 8.45)^2

x^2 + 182.85 = x^2 + 16.9x + 71.4025

182.25 = 16.9x + 71.4025

-16.9x = -110.8475

x = 6.559

Answer: x = 6.6 cm
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Suppose 6 quarts of a solution that is 52% antifreeze is mixed with 10 quarts of a solution that is 32% antifreeze. (A) how many
Vadim26 [7]
The total amount of the resulting mixture can be calculated by adding up the volume of the given substances assuming that volume addition is applicable given the properties of the fluids used. 

That is,
   T = 6 quarts + 10 quarts = 16 quarts

The total volume of the resulting mixture is 16.

Then, we do the component (antifreeze) balance by adding up the resulting antifreeze from the substances to the total. We let x be the percentage of antifreeze in the final mixture.

    6(0.52) + 10(0.32) = 16(x)

The value of x from the equation is 0.395.

Therefore, the answer to this item is 39.5%. 
4 0
3 years ago
Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -
GaryK [48]

Consider all parabolas:

1.

y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0
3 years ago
Im stuck on this question i need help
Korolek [52]
I think it’s one half 1/2
4 0
3 years ago
What kind of angle pair do they create?
Bond [772]

Answer:

They create supplementary angle pairs.

Step-by-step explanation:

Supplementary angles add up to 180°.

8 0
3 years ago
Help ?!-10+3x=8 <br> X=??
jekas [21]
Ok, so first off you need to remember that your goal is to get x alone.
How do you do this? Well, let's get rid of that pesky negative ten, shall we?
The opposite of a negative is positive, so you can move negative ten to the other side of the problem by adding ten. Make sure that you do operations on both sides. Once you add, you have 3x=18. When a number is right next door to an x, it means it is being multiplied. To get rid of the three, divide by three on both sides. Now your get x=6, which is your answer. Hopefully, I explained this well! :D
8 0
3 years ago
Read 2 more answers
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