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Jobisdone [24]
3 years ago
6

I don’t understand this. I can’t process any of it. Someone please help me I need to understand algebra 1

Mathematics
1 answer:
S_A_V [24]3 years ago
3 0

How can I help you with your algebra

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Find dy/dx using implicit differentiation<br> cos x + sin y = 1
777dan777 [17]

Differentiate both sides with respect to <em>x</em>, using the chain rule for the sine term:

\cos(x) + \sin(y) = 1

\implies -\sin(x) + \cos(y)\dfrac{\mathrm dy}{\mathrm dx} = 0

Solve for d<em>y</em>/d<em>x</em> :

-\sin(x) + \cos(y)\dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ \cos(y) \dfrac{\mathrm dy}{\mathrm dx} = \sin(x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\sin(x)}{\cos(y)}}

8 0
3 years ago
Solve 2y=8x 6 show it on graph
Umnica [9.8K]
2y = 8x + 6
y = 4x + 3
The graph is a straight line passing through points (0, 3) and (-3/4, 0) with slope = 4.
4 0
3 years ago
WILL GIVE BRAINLIEST!!!!
Vesna [10]

(2^8 \times 5^{-5} \times 19^0)^{-2} \times (\dfrac{5^{-2}}{2^3})^4 \times 2^{28} =

= 2^{-16} \times 5^{10} \times 1 \times \dfrac{5^{-8}}{2^{12}} \times 2^{28}

= 2^{-16} \times 5^{10} \times 5^{-8} \times 2^{-12} \times 2^{28}

= 2^{-16} \times 2^{-12} \times 2^{28} \times 5^{10} \times 5^{-8}

= 2^{-16-12+28} \times 5^{10-8}

= 2^{0} \times 5^{2}

= 1 \times 25

= 25

7 0
3 years ago
I need to know the answer
antoniya [11.8K]
2 days late, but the answer is no solution. solving a system of equations means finding where they intersect, but by looking at these equations, you know that they never intersect--they're parallel.

they share a slope (2), making them either parallel or "the same line", but the different x-intercepts (9 and -9) mean that they're different lines. they have no solution, or no intersection point, because they're parallel lines.
6 0
3 years ago
5. Find the general solution to y'''-y''+4y'-4y = 0
CaHeK987 [17]

For any equation,

a_ny^(n)+\dots+a_1y'+a_0y=0

assume solution of a form, e^{yt}

Which leads to,

(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0

Simplify to,

e^{yt}(y^3-y^2+4y-4)=0

Then find solutions,

\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}

For non repeated real root y, we have a form of,

y_1=c_1e^t

Following up,

For two non repeated complex roots y_2\neq y_3 where,

y_2=a+bi

and,

y_3=a-bi

the general solution has a form of,

y=e^{at}(c_2\cos(bt)+c_3\sin(bt))

Or in this case,

y=e^0(c_2\cos(2t)+c_3\sin(2t))

Now we just refine and get,

\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}

Hope this helps.

r3t40

5 0
3 years ago
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