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2 years ago

Which graph represents y=1/2x+2 please help!!

Mathematics

1 answer:

wel2 years ago

8 0

Answer:

Step-by-step explanation:

Let’s first look at the y-intercept. Since the equation is in y = mx + b form, we know that m is the slope and b is the y-intercept, so the slope is 1/2 and the y-intercept is 2, or (0,2). We can narrow down our search by noticing that C and D don’t intersect (0,2). That leaves A and B.

The slope is calculated by (y_1 - y_2)/(x_1 - x_2) given points (x_1, y_1) and (x_2, y_2). The slope for A is (4-0)/(2-(-2)) = 4/4 = 1 and the slope for B is (2-0)/(0-(-4)) = 2/4 = 1/2. The slope of B matches the slope that we are looking for. So, the answer is B.

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320/80 +3 = c

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2 years ago

My father's car travel on average 33.4miles on each gallon of petrol. How far will he travel on 55 gallons.

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2 years ago

Question 1 (1 point)

salantis [7]

Blank 1: 1

Blank 2: 3

Blank 3: -2

3 0

2 years ago

Hello pls help I'm dumb thx it's pre calc

Bumek [7]

(f-g)(x)=f(x)-g(x)=
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3 0

3 years ago

If f(x)=x^3-x+2, then (f^-1)'(2)

yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

$f\left(f^{-1}(x)\right) = x$

$\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x$

which is a cubic polynomial in $f^{-1}(x)$ with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

$\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}$

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0 or x = 1 or x = -1

Then $\left(f^{-1}\right)'(2)$ can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0

1 year ago