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2 years ago

Solve for x

}^{2} = 7x + 2" alt="4 {x}^{2} = 7x + 2" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Alina [70]2 years ago

7 0

$\bf 4x^2=7x+2\implies 4x^2-7x-2=0\implies (4x+1)(x-2)=0 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 4x\cdot x=\boxed{4x^2}~\hfill (1)(-2)=\boxed{-2}\hfill (4x\cdot -2)+(1\cdot x)=\boxed{-7x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 4x+1=0\implies 4x=-1\implies &x=-\cfrac{1}{4}\\[1em] x-2=0\implies &x=2 \end{cases}$

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3 years ago

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

Vinil7 [7]

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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

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1 year ago

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3 years ago

Read 2 more answers

-2×-7y=30 <br>7×+4y=18 <br>×= <br>y=

Sati [7]

Answer:

Answer is -7 Hope this helps :D

Step-by-step explanation:

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