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3 years ago

10

These are 3 questions regarding The Explicit Rule, and Arithmetic Sequence. They are worth 16 points. Thank you to any one who c

an help. :)

1 answer:

KiRa [710]3 years ago

3 0

That would be an = 4 + 3(n-1)

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What is constant of proportionality

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The constant of proportionality is the rate of growth. It is represented by y/x=k

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4 years ago

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Keaton spent half of his weekly allowance buying pizza. To earn more money his parents let him clean the windows in the house fo

Oksana_A [137]

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3 years ago

9z − 6z − –7z + –4 = 16 z=

Firdavs [7]

Answer:

z=- $\frac{2}{3}$

Step-by-step explanation:

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4 years ago

What is the standard form given the vertex (-3,3) and the focus point (-3,2)

baherus [9]

Answer:

$y=-\frac{1}{4}x^2-\frac{3}{2}x+\frac{3}{4}$

Step-by-step explanation:

The standard form of a parabola is written as

$y=ax^2+bx+c$

where a, b and c are the coefficients of the second-degree, first degree and zero-degree terms.

The coordinates of the vertex of a parabola is given by:

$x_b = -\frac{b}{2a}$

$y_b=c-\frac{b^2}{4a}$

The coordinates of the focus instead are given by

$x_f=-\frac{b}{2a}$

$y_f=y_b+\frac{1}{4a}$

In this problem, we know the coordinates of the vertex and of the focus point:

Vertex: (-3,3)

Focus point: (-3,2)

So we have:

$x_b=-3=-\frac{b}{2a}$ (1)

$y_b=3=c-\frac{b^2}{4a}$ (2)

$y_f=2=y_b+\frac{1}{4a}$ (3)

From eq.(1) we get

$2a=\frac{b}{3}$ (4)

Substituting into (2),

$3=c-\frac{b^2}{2(b/3)}\\3=c-\frac{3}{2}b\\c=3+\frac{3}{2}b$ (5)

Now rewriting eq.(3) as

$2=y_b+\frac{1}{4a}\\2=(c-\frac{3}{2}b)+\frac{1}{4a}$

And substituting (4) and (5) into this, we can find b:

$2=((3+\frac{3}{2}b)-\frac{3}{2}b)+\frac{1}{2(b/3)}\\2=3+\frac{3}{2b}\\-1=\frac{3}{2b}\\b=-\frac{3}{2}$

Then we can find a and c:

$2a=\frac{b}{3}\\a=\frac{b}{6}=\frac{-3/2}{6}=-\frac{1}{4}$

And

$c=3+\frac{3}{2}b=3+\frac{3}{2}(-\frac{3}{2})=3-\frac{9}{4}=\frac{3}{4}$

So the parabola is

$y=-\frac{1}{4}x^2-\frac{3}{2}x+\frac{3}{4}$

4 0

4 years ago

I can't figure out how to do (i + j) x (i x j)for vector calc

Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

$\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}$

Then, solving the determinant

$\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}$

In our case,

$\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}$

Where we used the formula for AxB to calculate ixj.

Finally,

$\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}$

Thus, (i+j)x(ixj)=i-j

8 0

1 year ago