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Rudiy27
3 years ago
10

Write a vector equation of the line that passes through P(4, 7) and is parallel to a = (3, 8).

Mathematics
1 answer:
love history [14]3 years ago
4 0

Answer:

B

Step-by-step explanation:

(x,y) = P + (a)t

(x,y) = (4,7) + t(3,8)

(x,y) - (4,7) = t(3,8)

(x-4, y-7) = t(3,8)

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I think the answer is 210.442
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Solve the equations in two different ways:6p=0.6(5p+15)
Damm [24]

Answer:

Step-by-step explanation:

1st way:

Solve directly

6p=0.6(5p+15)

Open bracket

6p = 3p + 9

Collecting like terms

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3p = 9

Dividing by 3

p = 9/3

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2nd way:

6p=0.6(5p+15)

Convert the decimal to fraction

6p= 6/10(5p+15)

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6p= 3/5(5p) +3/5(15)

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Dividing by 3

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6 0
4 years ago
Can y’all help me please? :))
german

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Step-by-step explanation:

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6 0
3 years ago
How many sides do 1 triangle, 4 octagons, 5 dodecagons, and 4 quadrilaterals have in all?
EleoNora [17]

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6 0
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

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

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

now, we can solve integral

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

=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|

now, we can plug bounds

and we get

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

so,

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

5 0
3 years ago
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