<h3>
Answer: y = x+1</h3>
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Explanation:
f(x) = x^3 - 2x + 3
f ' (x) = 3x^2 - 2 ..... apply the power rule
f ' (1) = 3(1)^2 - 2 ... plug in x coordinate of given point
f ' (1) = 1
If x = 1 is plugged into the derivative function, then we get the output 1. This means the slope of the tangent line at (1,2) is m = 1. It's just a coincidence that the x input value is the same as the slope m value.
Now apply point slope form to find the equation of the tangent line
y - y1 = m(x - x1)
y - 2 = 1(x - 1)
y - 2 = x - 1
y = x - 1 + 2
y = x + 1 is the equation of the tangent line.
The graph is shown below. I used GeoGebra to make the graph.
Answer:
its 1
Step-by-step explanation:
it literly says it on the grid and its really simple
Answer:
40
Step-by-step explanation:
remember PEMDAS
our equation looks like tihs: 
1. parentheses: in the denominator we have a parentese inside of another parenthese, so start with the inside first: (3-5) = -2.
so now our equation looks liek this 
now we have a mini PEMDAS situation within the parentheses: multiplication comes before addition/subtraction, so multiply -2*2 = -4, so now your equation looks like this: 
6-4 =2: so now we're done with parentheses part of PEMDAS, the only thing left is dividing 80/2 which equals 40.
Hello there!
Jenny wants to park in an airport parking lot. The parking lot costs x amount of money per hour, plus an additional parking fee of $10. Jenny stays in the parking lot for 3 hours.
This causes our equation to be 3x (3 hours) + 10 (additional parking fee).
I hope this helps!
Isolate the variable by dividing each side by factors that don't contain the variable.
a=- \frac{ x^{2}- \sqrt{( x^{3} + x^{2} b+12x-72(x-2)-2x} }{x-2} ,- \frac{ x^{2}+ \sqrt{( x^{3} + x^{2} b+12x-72(x-2)-2x} }{x-2}
Solve for b by simplifying both sides of the equation then isolating the variable.
b= \frac{12}{x}+ \frac{72}{ x^{2} }-2+2a- \frac{4a}{x}+ \frac{ a^{2} }{x}- \frac{2a^{z} }{ x^{2} }
Hopefully i helped ^.^ Mark brainly if possible. Lol once again i saw the same question so why not answer it again!
