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3 years ago

Let f(x)=4-x^2,find ( f(x+h)-f(x) ) / h and simplify

Mathematics

1 answer:

Alja [10]3 years ago

7 0

If $f(x)=4-x^2$ , then

$f(x+h)=4-(x+h)^2=4-(x^2+2xh+h^2)=4-x^2-2xh-h^2$

$\dfrac{f(x+h)-f(x)}h=\dfrac{(4-x^2-2xh-h^2)-(4-x^2)}h=-\dfrac{2xh+h^2}h$

Then for $h\neq0$ , we can cancel common terms:

$\dfrac{f(x+h)-f(x)}h=-2x+h$

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67 1/2 * (-1/3) it is due on Wednesday and for some reason I cont figure out this one

krok68 [10]

Answer:

-45/2

Step-by-step explanation:

67 1/2 * (-1/3)

67 1/2 = 135/2

So 135/2*1/3=45/2

45/2*-1=-45/2

7 0

1 year ago

4(4b-8)-3/8 Please help i have no idea what to do

Ksivusya [100]

Answer:

$4(4b - 8) - \frac{3}{8} \\ 16b - 32 - \frac{3}{4} \\ \frac{64b - 128 - 3}{8} \\ \frac{64b - 131}{8}$

hope this will help you

7 0

2 years ago

Which equation has a graph that is perpendicular to the graph of -x + 6y = -12?

serious [3.7K]

Answer:

c) 6x + y = -52 is required equation perpendicular to the given equation.

Step-by-step explanation:

If the equation is of the form : y = mx + C.

Here m = slope of the equation.

Two equations are said to be perpendicular if the product of their respective slopes is -1.

Here, equation 1 : -x + 6y = -12

or, 6y = -12 + x

or, y = (x/6) - 2

⇒Slope of line 1 = (1/6)

Now, for equation 2 to be perpendicular:

Check for each equation:

a. x + 6y = -67 ⇒ 6y = -67 - x

or, y = (-x/6) - (67/6) ⇒Slope of line 2 = (-1/6)

but $\frac{1}{6} \times \frac{-1}{6} \neq -1$

b. x - 6y = -52 ⇒ -6y = -52 - x

or, y = (x/6) + (52/6) ⇒Slope of line 2 = (1/6)

but $\frac{1}{6} \times \frac{1}{6} \neq -1$

c. 6x + y = -52

or, y =y = -52 - 6x ⇒Slope of line 2 = (-6)

$\frac{1}{6} \times (-6) = -1$

Hence, 6x + y = -52 is required equation 2.

d. 6x - y = 52 ⇒ -y = 52 - 6x

or, y = 6x - 52 ⇒Slope of line 2 = (6)

but $\frac{1}{6} \times 6 \neq -1$

Hence, 6x + y = -52 is the only required equation .

3 0

3 years ago

Read 2 more answers

12x-5y=-20 y=x+4 Please help me asap it's due Wednesday :(

Ahat [919]

Answer:

Point Form:

(0,4)

Equation Form:

x=0,y=4

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

5 0

3 years ago

Flying against the wind, an airplane travels 7520 kilometers in 8 hours. Flying with the wind, the same plane travels 6500 kilom

bezimeni [28]

Answer:

Rate of plane in still air = 1120 km/h

Rate of wind = 180 km/h

Step-by-step explanation:

Let the rate (speed) of the plane in still air = u km/h

Let the rate (speed) of the wind = v km/h

Formula for speed is:

$Speed = \dfrac{Distance}{Time}$

Given that against the wind, airplane travels 7520 km in 8 hours.

The speed against the wind will be slower so,

Against the wind, speed = (u -v ) km/h

Using above formula:

$(u-v) = \dfrac{7520}{8} km/h\\(u-v) = 940 km/h ..... (1)$

Given that with the wind, airplane travels 6500 km in 5 hours.

The speed with the wind will be faster so,

With the wind, speed = (u +v ) km/h

Using above formula:

$(u+v) = \dfrac{6520}{5} km/h\\(u+v) = 1300 km/h ..... (2)$

Adding (1) and (2) to solve for u and v:

$2u = 2240\\\Rightarrow u = 1120\ km/h$

Putting u in (1):

$1120-u=940\\\Rightarrow u = 180\ km/h$

So, the answers are:

Rate of plane in still air = 1120 km/h

Rate of wind = 180 km/h

6 0

3 years ago