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Ugo [173]
3 years ago
8

I need someone to teach me the tricks of linear equation​

Mathematics
2 answers:
Vladimir [108]3 years ago
6 0

Answer:

You can (a) use the old t-table approach; (b) you can draw a line and then figure out its slope and y-intercept; or you can (c) first explain the slope-intercept formula and then explain how the equation aligns with the formula.

OR.

The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it's pretty easy to find both intercepts (x and y).

Step-by-step explanation:

I hope this helped you gurl if not i apologize...

NikAS [45]3 years ago
4 0

Answer:

of what exactly and thanks for the points

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Suppose a tub has the shape of an elliptical paraboloid given by z = ax2+by2 (where a, b are some positive constants). If a marb
Umnica [9.8K]

Answer:

It would roll in this direction.

\nu  =  (-a/\sqrt{a^2+b^2},-b/\sqrt{a^2+b^2})

Step-by-step explanation:

It would roll to the direction of maximum decrease, which is the -1 times the direction of maximum increase, which is given by the gradient of the function.  

Since  

z =  ax^2  + by^2

For this case, the gradient of your function would be

\nabla z  = (2ax , 2by)

And  -1  times the gradient of your function would be

-\nabla z  = (-2ax , -2by)\\

Then, at

 (1,1,a+b),\\x = 1 \\y = 1

So it would go towards

v = (-2a,-2b)

The magnitud of that vector is

|v| =  2\sqrt{a^2+b^2}

and to conclude it would roll in this direction.

\nu  =  (-a/\sqrt{a^2+b^2},-b/\sqrt{a^2+b^2})

6 0
3 years ago
Find two consecutive odd integers whose sum is 116 (and please have steps) thank you
matrenka [14]
It's 57 and 59
Call (n) is the first odd interger => the second odd interger is (n+2)
=> n + (n + 2) = 116 
=> n + n + 2 = 116 
=> 2n = 114 
=> n = 57 
First odd interger is 57 and the second odd interger is 59 .
Sorry my English so bad =)))
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3 years ago
Given cosθ=√3/3 and sinθ<0. What is the value of sinθ?
KATRIN_1 [288]

\sin^{2} \theta+\cos^{2} \theta=1 \\ \\ \sin^{2} \theta+\left(\frac{\sqrt{3}}{3} \right)^{2}=1 \\ \\ \sin^{2} \theta+\frac{1}{3}=1 \\ \\ \sin^{2} \theta=\frac{2}{3} \\ \\ \theta=-\frac{\sqrt{2}}{\sqrt{3}}=\boxed{-\frac{\sqrt{6}}{3}}

5 0
2 years ago
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