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3 years ago

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In ΔIJK, the measure of ∠K=90°, KJ = 63, IK = 16, and JI = 65. What ratio represents the sine of ∠I?

1 answer:

FinnZ [79.3K]3 years ago

5 0

The sine of ∠I is represented by the ratio 63:65

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Answer:

See below

Step-by-step explanation:

Since you didn't give an actual quadratic equation and just the standard form equation, I will reduce ax^2+bx+c=0:

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Find an equation of the plane with the given characteristics. The plane passes through the points (4, 3, 1) and (4, 1, -7) and i

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Answer:

$3x - 4y + z = 1$

Step-by-step explanation:

Given

$Point\ 1 = (4,3,1)$

$Point\ 2 = (4,1,-7)$

Perpendicular to $8x + 7y + 4z = 18$

Required

Determine the plane equation

The general equation of a plane is:

$a(x-x_1) + b(y - y_1) + c(z-z_1) = 0$

For $n =$

$(x_1,y_1,z_1) = (4,3,1)$

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First, we need to determine parallel vector $V_1$

$V_1 =$

$V_1 =$

$V_1 =$

$V_1$ is parallel to the required plane

From the question, the required plane is perpendicular to $8x + 7y + 4z = 18$

Next, we determine vector $V_2$

$V_2 =$

This implies that the required plane is parallel to $V_2$

Hence: $V_1$ and $V_2$ are parallel.

So, we can calculate the cross product $V_1 * V_2$

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$V_2 =$

$n = V_1 * V_2$

$V_1 * V_2 =\left[\begin{array}{ccc}i&j&k\\0&2&8\\8&7&4\end{array}\right]$

The product is always of the form + - +

So:

$V_1 * V_2 = i\left[\begin{array}{cc}2&8\\7&4\end{array}\right]$ $-j\left[\begin{array}{cc}0&8\\8&4\end{array}\right]$ $+k\left[\begin{array}{cc}0&2\\8&7\end{array}\right]$

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$V_1 * V_2 = i(2*4- 8*7) - j(0*4- 8*8) + k(0*7 - 2 * 8)$

$V_1 * V_2 = i(8- 56) - j(0- 64) + k(0 - 16)$

$V_1 * V_2 = i(-48) - j(- 64) + k(- 16)$

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So, the resulting vector, n is:

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Recall that:

$n =$

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Substitute these values in $a(x-x_1) + b(y - y_1) + c(z-z_1) = 0$

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Collect Like Terms

$-48x + 64y - 16z = 0 - 192 + 192 - 16$

$-48x + 64y - 16z = -16$

Divide through by -16

$3x - 4y + z = 1$

<em>Hence, the equation of the plane is</em> $3x - 4y + z = 1$ <em></em>

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3 years ago

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Answer:

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3 years ago

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Answer:

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3 years ago