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Vlad [161]
3 years ago
7

What is the slope of the line? 7x+2y=57x+2y=5

Mathematics
1 answer:
andre [41]3 years ago
6 0

1/10=x hope that helps a bunch

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Which subject did most students choose? Explain
Snezhnost [94]
English...because english subject is a part of our language worldwide and that can communicate all over the world...its just my opinion
8 0
3 years ago
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Solve the system of equations 4x+5y=-1 and -5x-8y=10 by combining the equations.
Nimfa-mama [501]

The solution of the equation are as follows:

x = 6 and y = -5

<h3 />

<h3>How to solve the system of equation?</h3>

4x + 5y = -1

-5x - 8y = 10

Therefore,

20x + 25y = -5

-20x - 32y = 40

-7y = 35

y = -5

Hence,

4x + 5(-5) = -1

4x - 25 = -1

4x = -1 + 25

4x = 24

x = 24 / 4

x = 6

learn more on equation here:brainly.com/question/14383993

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6 0
1 year ago
How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
Can you help me please?
Artist 52 [7]

Answer:

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
1. What is the slope of the line tangent to the curve defined by y2 + xy - x2 = 11x at the point (2, 3)?
andreyandreev [35.5K]

Step-by-step explanation:

By using Implicit Differentiation,

d/dx (y² + xy - x²) = d/dx (11x)

d/dx (y²) + d/dx (xy) - d/dx (x²) = 11

2y * dy/dx + x * dy/dx + y - 2x = 11

dy/dx (2y + x) = 11 + 2x - y

dy/dx = (11 + 2x - y) / (2y + x).

At the point (2, 3), we have x = 2, y = 3.

=> dy/dx = (11 + 2(2) - (3)) / (2(3) + (2))

= 12 / 8 = 1.5.

P.S. Your question is weird because (2,3) is not on the graph, let me know what is the correct question thanks!

7 0
2 years ago
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