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3 years ago

PLEASE GUYS HELP ITS DUE TONIGHT

Mathematics

1 answer:

KATRIN_1 [288]3 years ago

4 0

Answer:

20. AB = 42

21. BC = 28

22. AC = 70

23. BC = 20.4

24. FH = 48

25. DE = 10, EF = 10, DF = 20

Step-by-step explanation:

✍️Given:

AB = 2x + 7

BC = 28

AC = 4x,

20. Assuming B is between A and C, thus:

AB + BC = AC (Segment Addition Postulate)

2x + 7 + 28 = 4x (substitution)

Collect like terms

2x + 35 = 4x

35 = 4x - 2x

35 = 2x

Divide both side by 2

17.5 = x

AB = 2x + 7

Plug in the value of x

AB = 2(17.5) + 7 = 42

21. BC = 28 (given)

22. AC = 4x

Plug in the value of x

AC = 4(17.5) = 70

✍️Given:

AC = 35 and AB = 14.6.

Assuming B is between A and C, thus:

23. AB + BC = AC (Segment Addition Postulate)

14.6 + BC = 35 (Substitution)

Subtract 14.6 from each side

BC = 35 - 14.6

BC = 20.4

24. FH = 7x + 6

FG = 4x

GH = 24

FG + GH = FH (Segment Addition Postulate)

$4x + 24 = 7x + 6$ (substitution)

Collect like terms

$4x - 7x = -24 + 6$

$-3x = - 18$

Divide both sides by -3

$x = 6$

FH = 7x + 6

Plug in the value of x

FH = 7(6) + 6 = 48

25. DE = 5x, EF = 3x + 4

Given that E bisects DF, therefore,

DE = EF

5x = 3x + 4 (substitution)

Subtract 3x from each side

5x - 3x = 4

2x = 4

Divide both sides by 2

x = 2

DE = 5x

Plug in the value of x

DE = 5(2) = 10

EF = 3x + 4

Plug in the value of x

EF = 3(2) + 4 = 10

DF = DE + EF

DE = 10 + 10 (substitution)

DE = 20

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Step-by-step explanation:

Given parametric equations are x = t + sin(t) and y = cos(t) and given interval is

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$\int\limits^a_b {\sqrt{\frac{(dx}{dt}) ^{2}+(\frac{dy}{dt}) ^2 } } \, dx$

Given limits values are −π ≤ t ≤ π

x = t + sin(t) ...….. (1)

y = cos(t).......(2)

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differentiating equation (2) with respective to 'y'

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The length of curve is

$\int\limits^\pi_\pi {\sqrt{(1+cost)^{2}+(-sint)^2 } } \, dt$

$\int\limits^\pi_\pi \, {\sqrt{(1+cost)^{2}+2cost+(sint)^2 } } \, dt$

on simplification , we get

here using sin^2(t) +cos^2(t) =1 and after simplification , we get

$\int\limits^\pi_\pi \, {\sqrt{(2+2cost } } \, dt$

$\sqrt{2} \int\limits^\pi_\pi \, {\sqrt{(1+1cost } } \, dt$

again using formula, 1+cost = 2cos^2(t/2)

$\sqrt{2} \int\limits^\pi _\pi {\sqrt{2cos^2\frac{t}{2} } } \, dt$

Taking common $\sqrt{2}$ we get ,

$\sqrt{2}\sqrt{2} \int\limits^\pi _\pi ( {\sqrt{cos^2\frac{t}{2} } } \, dt$

$2(\int\limits^\pi _\pi {cos\frac{t}{2} } \, dt$

$2(\frac{sin(\frac{t}{2} }{\frac{t}{2} } )^{\pi } _{-\pi }$

length of curve = $4(sin(\frac{\pi }{2} )- sin(\frac{-\pi }{2} ))$

length of the curve is = 4(1+1) = 8

<u>conclusion</u>:-

The arrow of the direction or the length of curve = 8

7 0

3 years ago

PLEASE GUYS HELP ITS DUE TONIGHT​

PLEASE GUYS HELP ITS DUE TONIGHT