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3 years ago

Y=2x-3 -- How do you graph this

Mathematics

1 answer:

Debora [2.8K]3 years ago

8 0

Answer:

y = 2x - 3

To plot a graph

you need at least one coordinate and slope

2 coordinates

y = 2x -3

By Comparing with General Equation of a line that is y = mx + c

Here m represents slope

and c is y intercept

which inturn results in m = 2 , c = -3

c = -3

So Coordinate is (0,-3)

Plot (0,-3)

Take a line having slope 2

Then you can plot it

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p = -1 and q = 10
answer is A.

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3 years ago

System OT<br> W + b =13<br> 6.5w + 2b = 57.5

mixer [17]

Answer: w = 7, x = 6

Step-by-step explanation: Solve by substitution

W + b = 13

rewrite as b = 13 - w and substitute that value for b in the second equation

6.5w + 2b = 57.5 Then solve for w

6.5w + 2(13-w) = 57.5 . Distribute

6.5w + 26 - 2w = 57.5 . Subtract 26 from both sides. Combine like terms and simplify

6.5w - 2w = 57.5 - 26

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3 years ago

Find dy/dx by implicit differentiation.

kow [346]

dy/dx by implicit differentiation is cos(πx)/sin(πy)

<h3>How to find dy/dx by implicit differentiation?</h3>

Since we have the equation

(sin(πx) + cos(πy)⁸ = 17, to find dy/dx, we differentiate implicitly.

So, [(sin(πx) + cos(πy)⁸ = 17]

d[(sin(πx) + cos(πy)⁸]/dx = d17/dx

d[(sin(πx) + cos(πy)⁸]/dx = 0

Let sin(πx) + cos(πy) = u

So, du⁸/dx = 0

du⁸/du × du/dx = 0

Since,

du⁸/du = 8u⁷ and

du/dx = d[sin(πx) + cos(πy)]/dx

= dsin(πx)/dx + dcos(πy)/dx

= dsin(πx)/dx + (dcos(πy)/dy × dy/dx)

= πcos(πx) - πsin(πy) × dy/dx

So, du⁸/dx = 0

du⁸/du × du/dx = 0

8u⁷ × [ πcos(πx) - πsin(πy) × dy/dx] = 0

8[(sin(πx) + cos(πy)]⁷ × (πcos(πx) - πsin(πy) × dy/dx) = 0

Since 8[(sin(πx) + cos(πy)]⁷ ≠ 0

(πcos(πx) - πsin(πy) × dy/dx) = 0

πcos(πx) = πsin(πy) × dy/dx

dy/dx = πcos(πx)/πsin(πy)

dy/dx = cos(πx)/sin(πy)

So, dy/dx by implicit differentiation is cos(πx)/sin(πy)

Learn more about implicit differentiation here:

brainly.com/question/25081524

#SPJ1

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2 years ago