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3 years ago

Negative 7 divided by 14/3 as a fraction

Mathematics

2 answers:

masya89 [10]3 years ago

4 0

I will answer your question. It is -3/2

Oliga [24]3 years ago

3 0

If you divide -7 by 14/3 you get -3/2

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Find the y intercept

erica [24]

Step-by-step explanation:

Slope of line ST = Slope of line PQ = -3.

y = -3x + c

When x = 3, y = -1

=> (-1) = -3(3) + c, c = 8

Therefore the equation of line ST is y = -3x + 8 and the y-intercept is 8.

7 0

3 years ago

Read 2 more answers

Can an eulerian circuit also be an eulerian path?

kap26 [50]

Answer: i think yes

Step-by-step explanation:

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2 years ago

My teacher didn’t really teach this and um how do I do this

Lorico [155]

Answer:

Step-by-step explanation:

1. 3.8 x 10^5 + 5.5 x 10^5 = 930,000 or 9.3 x 10^5

You just have to solve the equation and shorten the answer. For number one the answer is 930,000, so you have to make it into a number with decimals since a scientific notation number can not be equal to 10 or be greater. So you have to round the number to 9.3 not 93. To "round" it to 9.3 you have to move the decimal at the end of 930,000 to be in between the numbers 9 and 3. How many times you move the decimal to the left or right, that is the exponent of 10. So for 930,000 you moved the decimal 5 times to the left so the exponent for 10 is positive 5.

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3 years ago

Simplify the following radical <br> Sqrt 126x^13

fenix001 [56]

Answer:

$\large\boxed{\sqrt{126x^{13}}=3x^6\sqrt{14x}}$

Step-by-step explanation:

$Domain:\ x\geq0\\\\\sqrt{126x^{13}}=\sqrt{9\cdot14\cdot x^{12+1}}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{9\cdot14\cdot x^{12}\cdot x^1}=\sqrt{9\cdot14\cdot x^{6\cdot2}\cdot x}\\\\\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\ \text{and}\ (a^n)^m=a^{nm}\\\\=\sqrt9\cdot\sqrt{14}\cdot\sqrt{(x^6)^2}\cdot\sqrt{x}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=3\cdot\sqrt{14}\cdot x^6\cdot\sqrt{x}=3x^6\sqrt{14x}$

3 0

3 years ago

Find the equation of the linear function represented by the table below in slope-

7nadin3 [17]

to get the equation of any straight line we simply need two points off of it, hmm let's get two from this table

$\begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} 0&4\\ 1&10\\ 2&16\\ 3&22\\ 4&28\\ \cline{1-2} \end{array} \begin{array}{llll} \\ \leftarrow \textit{let's use this point}\\\\ \leftarrow \textit{and this point} \end{array}$

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{22}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{22}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}}\implies \cfrac{12}{2}\implies 6 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{6}(x-\stackrel{x_1}{1}) \\\\\\ y-10=6x-6\implies y=6x+4$

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1 year ago