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mel-nik [20]
2 years ago
5

Complemetary angles are two angles whose measures add up to 90 degrees. The ratio of the measures of two complementary angles is

2:13. What are the measures of the angles?
Mathematics
1 answer:
pochemuha2 years ago
3 0

Let A and B be the two complementary angles.

A = smaller angle = 2x

B = larger angle = 13x

x = some unknown number

Note how the ratio A:B turns into 2x:13x which simplifies to 2:13

A+B = 90 ... because the angles are complementary

2x+13x = 90 ... substitution

15x = 90

x = 90/15

x = 6

A = 2*x = 2*6 = 12 degrees

B = 13*x = 13*6 = 78 degrees

The two angles are 12 degrees and 78 degrees.

Check:

A/B = 12/78 = (2*6)/(13*6) = 2/13, so A:B = 2:13

A+B = 12+78 = 90

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GalinKa [24]

Answer:

x=7 and y=-1

Step-by-step explanation:

X+Y=6 OR X=6-Y  ...(1)

X-Y=8    ...(2)

substitue X=6-Y in (2)

(6-Y)-Y=8

6-2Y=8

-2Y=8-6

-2Y=2

Y=2/-2\Y=-1 ANS.

for x, substitute Y=-1 in (1) above

X-(-1)=8

X=8-1

X=7 ANS.

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2 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.

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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.

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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.

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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.

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i use the two sided t-test.

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generally in statistics, we use a two sided t-test when an area of a distribution is two-sided and tests whether a particular given sample in question  is greater than or less than a exact range of values.

the two sided t-test  is mostly used in null-hypothesis testing and testing for statistical significance.

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