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PSYCHO15rus [73]
3 years ago
5

Write the standard form of the line that passes through the point (1,5) and is parallel to the x -axis. Include your work in you

r final answer. Type your answer in the box provided to submit your solution.
Mathematics
1 answer:
scoray [572]3 years ago
7 0

Answer:

y = 5

Step-by-step explanation:

A line parallel to the x- axis has equation

y = c

Where c is the value of the y- coordinates the line passes through

The line passes through (1, 5) with y- coordinate of 5, thus

y = 5 ← equation of line

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State the largest solution of (x + 3)2 - 4 = 0.
Zinaida [17]
Hello :
(x+3)² -2² = 0
by identite :  a²-b² =(a-b)(a+b)
(x+3-2)(x+3+2)= 0
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3 years ago
Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4
arsen [322]

we are given

R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k

now, we can find x , y and z components

x=cos(8t),y=sin(8t),z=8ln(cos(t))

Arc length calculation:

we can use formula

L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt

x'=-8sin(8t),y=8cos(8t),z=-8tan(t)

now, we can plug these values

L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt

now, we can simplify it

L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt

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L=\int _0^{\frac{\pi }{4}}8sec(t) dt

now, we can solve integral

\int \:8\sec \left(t\right)dt

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now, we can plug bounds

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=8\ln \left(\sqrt{2}+1\right)-0

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L=8\ln \left(1+\sqrt{2}\right)..............Answer

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3 years ago
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Zinaida [17]

Answer:

60°

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