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Lerok [7]
3 years ago
8

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Line A passes t

hrough points (-1, 5) and (3, -1). Line B passes through points (7, 2) and (6, -1). At what point does line A intersect line B?
Mathematics
1 answer:
natulia [17]3 years ago
5 0
Line A
slope = (-1 - 5)/(3+1) = -6/4 = -3/2
y = mx + b
5 = -3/2(-1) + b
b = 5 - 3/2
b = 7/2
equation of line A: y  = -3/2x + 7/2

Line B
slope = (-1 - 2)/(6-7) = -3/-1 = 3
y = mx + b
2 = 3(7) + b
b = 2- 21
b = - 19
equation of line B: y  = 3x -19

line A intersect line B when -3/2x + 7/2 =  3x -19
so
-3/2x + 7/2 =  3x -19
3x + 3/2x = 19 + 7/2
9/2x = 45/2
     x = (45/2) * (2/9)
     x = 5

     y  = 3x - 19 = 3(5) -19 = 15 -19 = -4

    (5 , -4)

answer
line A intersect line B at (5, -4)
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T = 7/8 × 4  which equals  7/2


7/2 is your answer
4 0
2 years ago
Please help w this! Its a calculus question! look at the picture for the problem,
neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines x=\pm3\sqrt2 and y=\pm3\sqrt2.

Let R be the region bounded by the line x=3\sqrt2 and the circle x^2+y^2=36 (the rightmost blue region). The right side of the circle can be expressed in terms of x as a function of y:

x^2+y^2=36\implies x=\sqrt{36-y^2}

Then the area of this circular segment is

\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy

=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy

Substitute y=6\sin t, so that \mathrm dy=6\cos t\,\mathrm dt

=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt

=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18

Then the area of the entire blue region is 4 times this, a total of \boxed{36\pi-72}.

Alternatively, you can compute the area of R in polar coordinates. The line x=3\sqrt2 becomes r=3\sqrt2\sec\theta, while the circle is given by r=6. The two curves intersect at \theta=\pm\dfrac\pi4, so that

\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta

=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18

so again the total area would be 36\pi-72.

Or you can omit using calculus altogether and rely on some basic geometric facts. The region R is a circular segment subtended by a central angle of \dfrac\pi2 radians. Then its area is

\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18

so the total area is, once again, 36\pi-72.

An even simpler way is to subtract the area of the square from the area of the circle.

\pi6^2-(6\sqrt2)^2=36\pi-72

6 0
3 years ago
5.
Alex Ar [27]
Answer i believe is D!
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3 years ago
Your check register shows a balance of $565. Your last bank statement shows an ending balance of $715. All transactions match ex
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My opinion is answer A
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2 years ago
Complete the proof. Help. Please.
vladimir2022 [97]
The reason for 2 is simply dividing both sides of the expression in 1 by BE*BD. In 3, angle ABC=DBE because they are vertical angles formed by two intersecting lines (AE and CD). In 3, with these two conditions, we can conclude that the two triangles are similar by SAS. 
4 0
2 years ago
Read 2 more answers
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