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Neporo4naja [7]
3 years ago
10

Find the missing values for the exponential function represented by the table below.

Mathematics
1 answer:
postnew [5]3 years ago
8 0

Answer:

d. 13.5, 20.25

Step-by-step explanation:

An exponential function is monotonic and does not change sign. On this basis alone, the first three choices can be eliminated.

The common ratio is 6/4 = 3/2.

The term in the sequence after 9 is 9·(3/2) = 27/2 = 13.5.

The term in the sequence after 27/2 is (27/2)·(3/2) = 81/4 = 20.25.

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Solve the simultaneous equations 7 x + 4 y = 26 5 x + 4 y = 14
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Step-by-step explanation:

i used elimination method

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Which expression in equivalent to 12(2n+6) ?
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2 years ago
3x to the power of 3 + 21x squared + 36x = 0 by factoring the quadratic equation.
Diano4ka-milaya [45]

Answer:

x = 0, -4 and -3.

Step-by-step explanation:

The given expression is "3x to the power of 3 + 21x squared + 36x = 0".

We can write this expression as follows :

3x^3+21x^2+36x=0

Taking 3x common.

3x(x^2+7x+12)=0\\\\3x=0\ \text{and}\ (x^2+7x+12)=0

x = 3

x^2+7x+12=0\\\\x^2+3x+4x+12=0\\\\x(x+3)+4(x+3)=0\\\\(x+4)(x+3)=0\\\\x=-4,-3

Hence, the solution of the given quadratic equation are x = 0, -4 and -3.

5 0
3 years ago
Find two unit vectors orthogonal to both given vectors. i j k, 4i k
Maurinko [17]
The cross product of two vectors gives a third vector \mathbf v that is orthogonal to the first two.

\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k

Normalize this vector by dividing it by its norm:

\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)

To get another vector orthogonal to the first two, you can just change the sign and use -\mathbf v.
6 0
3 years ago
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