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olga_2 [115]
3 years ago
11

Please helppppppppppp​

Mathematics
2 answers:
Eduardwww [97]3 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

Substitute the given values of x into f : x → \frac{3}{7} x + 4

x = - 14 → (\frac{3}{7} × - 14) + 4 = (3 × - 2) + 4 = - 6 + 4 = - 2

x = 28 → ( \frac{3}{7} × 28) + 4 = (3 × 4) + 4 = 12 + 4 = 16

x = \frac{7}{8} → (\frac{3}{7} × \frac{7}{8} ) + 4 = \frac{3}{8} + 4 = \frac{3}{8} + \frac{32}{8} = \frac{35}{8}

x = - \frac{2}{9} → (\frac{3}{7} × - \frac{2}{9} ) + 4 = (\frac{1}{7} × - \frac{2}{3}) + 4 = - \frac{2}{21} + \frac{84}{21} = \frac{82}{21}

Hatshy [7]3 years ago
6 0

Answer:

See below in bold.

Step-by-step explanation:

1. 3/7 * -14 + 4

   = -6 + 4

    = -2.

2. 3/7 * 28 + 4

   = 12 + 4

    = 16.

3 3/7 * 7/8 + 4

   =  3/8 + 4

    =   35/8 or 4 3/8.

4. 3/7 * -2/9 + 4

   = -6/63 + 4

   = -2/21 + 4

    = -82/21 or -3 19/21.

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HELP! Find v × w if v = 3i + 8j – 6k and w = –4i – 2j – 3k.
wolverine [178]

Using the determinant method, the cross product is

\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\3&8&-6\\-4&-2&-3\end{vmatrix}=\begin{vmatrix}8&-6\\-2&-3end{vmatrix}\,\vec\imath-\begin{vmatrix}3&-6\\-4&-3\end{vmatrix}\,\vec\jmath+\begin{vmatrix}3&8\\-4&-2\end{vmatrix}\,\vec k=-36\,\vec\imath+33\,\vec\jmath+26\,\vec k

so the answer is B.

Or you can apply the properties of the cross product. By distributivity, we have

(3i + 8j - 6k) x (-4i - 2j - 3k)

= -12(i x i) - 32(j x i) + 24(k x i) - 6(i x j) - 16(j x j) + 12(k x j) - 9(i x k) - 24(j x k) + 18(k x k)

Now recall that

  • (i x i) = (j x j) = (k x k) = 0 (the zero vector)
  • (i x j) = k
  • (j x k) = i
  • (k x i) = j
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Putting these rules together, we get

(3i + 8j - 6k) x (-4i - 2j - 3k)

= -32(-k) + 24j - 6k + 12(-i) - 9(-j) - 24i

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