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sergejj [24]
3 years ago
11

New Saving

Mathematics
1 answer:
adelina 88 [10]3 years ago
4 0

Answer:

And i opssskkkkkkkkkkkk

Step-by-step explanation:

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Write in point slope form of the line that passes through the points (3,-1) and (6,-2)
skelet666 [1.2K]

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-2}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-2-(-1)}{6-3}\implies \cfrac{-2+1}{3}\implies -\cfrac{1}{3} \\\\\\ \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-(-1)=-\cfrac{1}{3}(x-3)\implies y+1=-\cfrac{1}{3}(x-3)

8 0
3 years ago
Please help <br><br> Please help<br><br> Please help
san4es73 [151]

Answer:

angle 1 = 52°

angle 2 = 90°

angle 3 = 38°

angle 4 = 38°

4 0
2 years ago
HELP ME PLEASE I REALLY NEED HELP!? :(
Aleksandr [31]
Just answers since long and many simplifiction
it's like doing those multiplication memory things that had like 100 questions but now with algebra fractions

5. fourth or \frac{5x-15}{x^{2}-3x-28}
6. third or\frac{4x+25}{x^{2}+5x+4}
7.the answe ris x+3 since it can be factored out of the 5x+15 tingie (second )
8. fourth or \frac{x^{2}+4x}{4}
9. fourth or \frac{2y+6}{9y-7} times \frac{3}{2}
10. third or \frac{(x+3)(x+8)}{(x-8)(x-8)}
11. second or \frac{x-5}{x-4}
12. third or 11(x+7)=60
5 0
3 years ago
What is the velocity of a car moving 20 m per second
murzikaleks [220]
20mps. Because velocity is the speed per unit. In this case it is going 20 m per second
5 0
3 years ago
Use Newton's method with initial approximation x1 = 1 to find x2, the second approximation to the root of the equation x4 − x −
drek231 [11]

Answer:

x_{2} = 0.0000

Step-by-step explanation:

The formula for the Newton's method is:

x_{i+1} = x_{i} + \frac{f(x_{i})}{f'(x_{i})}

Where f' (x_{i}) is the first derivative of the function evaluated in x_{i}.

x_{i+1} = x_{i} + \frac{x_{i}^{4}-x_{i}-3}{4\cdot x_{i}^{3}-1}

Lastly, the value of x_{2} is determined by replacing x_{1} with its numerical value:

x_{2} = x_{1} + \frac{x_{1}^{4}-x_{1}-3}{4\cdot x_{1}^{3}-1}

x_{2} = 1.0000 + \frac{1.0000^{4}-1.0000-3}{4\cdot (1.0000)^{3}-1}

x_{2} = 0.0000

8 0
3 years ago
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