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kramer
3 years ago
8

The product of 5 and the cube of x, increase by the difference of 6 and x3

Mathematics
1 answer:
lawyer [7]3 years ago
5 0

Answer:

4x^{3} +6

Step-by-step explanation:

  • The product of 5 and the cube of x

5*x^3\\\\=5x^3

  • the difference of 6 and x^3

6-x^3

I understand as increase as sum so

(5x^3)+(6-x^3)\\\\5x^3+6-x^3\\\\(5-1)x^3+6\\\\4x^3+6

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Can someone help me with 1& 2 will mark brainiest
melamori03 [73]
Sin 45° = opposite side / hypotenuse

1/2 = 8/h
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2)6/12 = 1/2
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6√3/12 = √3/2
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Sum of angles of triangle = 180°
The 3rd angle = 180 - (45 + 30) = 180 - 75 = 105°

Angles = 30° , 45° , 105°


7 0
3 years ago
What inverse operation is used to solve the equation? x 2 = 10
Alla [95]

Answer:

5

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? x 2 = 10

     /2     /2

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6 0
3 years ago
Read 2 more answers
F(1)=-16 f(n)=-29-f(n-1) f(2)=
charle [14.2K]
F(2)=-29-f(1) = -29-(-16)=-13
4 0
3 years ago
Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
Since each vector is a member of \mathbb R^3, the vectors will span \mathbb R^3 if they form a basis for \mathbb R^3, which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
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3 years ago
What is the slope of the line?
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Y sub 2- y sub 1 over x sub 2- x sub 1
4 0
3 years ago
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