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tekilochka [14]
3 years ago
15

The line to the right goes through the point (6, y). If the angle the graph makes with the x-axis is 22°, what is

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
8 0

Answer:

The y-coordinate of the point is approximately 2.42.

Step-by-step explanation:

From the figure attached below the statement we derive the following equation to calculate the y-coordinate (y), dimensionless, of the point in terms of the x-coordinate (x), dimensionless, and angle of the line (\theta), measured in sexagesimal degrees:

y = x\cdot \tan \theta (1)

If we know that x = 6 and \theta = 22^{\circ}, then the y-coordinate of the point is:

y = 6\cdot \tan 22^{\circ}

y\approx 2.42

The y-coordinate of the point is approximately 2.42.

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Answer:

Option B 0.04803

Step-by-step explanation:

we have

(4*\frac{1}{100})+(8*\frac{1}{1,000})+(3*\frac{1}{100,000})\\ \\ =(\frac{4}{100})+(\frac{8}{1,000})+(\frac{3}{100,000}) \\ \\ =0.04+0.008+0.00003\\ \\=0.04803

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You then add -8.5 and -8, because the device has lowered 8.5 degrees and 8 degrees. This equals -16.5 degrees, or a decrease of 16.5 degrees.
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Type the correct answer in each box. Use numerals instead of words.
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Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17   eq(1)

y = -1/2x            eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a = 5/4, b =0 and c = -17

x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}

Finding value of y:

y = -1/2x

y=-1/2(\frac{\pm2\sqrt{85}}{5})

y=\frac{\pm\sqrt{85}}{5}

System A has 4 real solutions.

System B

y = x^2 -7x + 10    eq(1)

y = -6x + 5            eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

a= 1, b =-1 and c =5

x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9    eq(1)

8x - y = -17        eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

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2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

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8(-2) - y = -17    

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-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

4 0
3 years ago
Read 2 more answers
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