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Anettt [7]
3 years ago
9

Use the product rule to simplify the radical— square root 315

Mathematics
1 answer:
Luden [163]3 years ago
5 0
  • Product rule with radicals: √ab = √a × √b

Using the product rule, simplify the radical as such:

√315 = √7 × √45 = √7 × √9 × √5 = 3 × √7 × √5 = 3√35

<u>The simplified radical of √315 is 3√35.</u>

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Which Part of the Angle is the Vertex?
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D  Point A

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The vertex is where the two sides of the angles come together

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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}

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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.

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\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
I need help! I don't know how to get the answer or what it is. Can anyone help me it's for a test and I'm on my last attempt?
Greeley [361]

Answer:

See explanation

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(4)

Using the sine ratio in  the right triangle

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(5)

Using the tangent ratio in the right triangle

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7 0
3 years ago
SOMEOJE HELLP PLSSSSSS
Neporo4naja [7]

Answer:

Step-by-step explanation:

this seems like substitution

If so its the first part is 12x6

second part is 35 x 6

third part is 56 x 6

7 0
3 years ago
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