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anzhelika [568]
3 years ago
7

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

So

\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<br> 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<br> y=x+4<br> Please help me asap it's due Wednesday<br> :(
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:

<em>Rate of plane in still air = 1120 km/h</em>

<em>Rate of wind = 180 km/h</em>

<em></em>

6 0
3 years ago
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