PLEASEE HELP MEEE<br> I need help with my math homework

1. ) Consider the function f(x)=5−7x2,−5≤x≤1

Marat540 [252]

1.) The interval of the value of x is from -5 to 1, inclusive. Remember that what is asked is the absolute value, thus the sign does not matter even if you have to subtract x from 5. Thus, the maximum value would be obtained if the x is smaller, which is 1. The minimum value is obtained when x=-5.

Absolute maximum value: x = - 5
f(-5) = ║5 - 7(-5)^2║ = ║-170║=170

Absolute minimum value: x = 1
f(1) = ║5 - 7(1)^2║ = ║-2║= 2

2.) The Mean Value Theorem (MVT) applies to functions that are continuous and differentiable on the closed and open interval of a to b, respectively. Since the function is a quadratic function, MVT can be applied. Then, this means that there is a value of c which is between a and b. This could be determined using this formula according to MVT:

$f'(c)= \frac{f(b)-f(a)}{b-a}$

The differentiated form would be f'(x) = -2x. Then,

$-2c = \frac{(4- 0^{2} )-(4- (-1)^{2}) }{0--1}=1$

$c=- \frac{1}{2}$

Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.

3 0

4 years ago

Cordell Johnson purchased a snow blower for $237.00 and two bags of rock salt for

skelet666 [1.2K]

Answer:

a.240.15

Step-by-step explanation:

a.240.15

b.12.0075

c.252.1575

3 0

2 years ago

Implicit differentiation of 1/x +1/y=5 y(4)= 4/19<br> y'(4)=?

Aneli [31]

$\frac{1}{x} +\frac{1}{y} = 5\\\\x^{-1}+y^{-1}=5\\$

Above, I changed the fraction form of x and y into exponential form so it is easier to see the differentiation. Now, we can differentiate:

$-1x^{-2}+-1y^{-2}\frac{dy}{dx}=5\\\\\frac{-1}{x^2}-\frac{1}{y^2}\frac{dy}{dx}=5\\\\-\frac{1}{y^2}\frac{dy}{dx}=5+\frac{1}{x^2}\\\\\frac{dy}{dx}=-5y^2-\frac{y^2}{x^2}$

Now that we have dy/dx, we can plug in the x, which is 4, and the y, which is 4/19. We know these values of x and y because your question stated y(4) = 4/19.

$\frac{dy}{dx}=-5(\frac{4}{19})^2-\frac{(\frac{4}{19})^2}{(4)^2}\\\\\frac{dy}{dx}=-5(\frac{16}{361})-\frac{(\frac{16}{361})}{16}\\\\\frac{dy}{dx}=\frac{-80}{361}-\frac{1}{361}\\\\\frac{dy}{dx}=\frac{-81}{361}$

5 0

4 years ago

A Ferrari drove 300 kilometers in one and a half hours. How fast was the car going?

irga5000 [103]

I'm assuming you mean how many mph when you refer to how fast the car was going.

So 300 km = 1.5 hrs

1.) Find the number of km per hour
Since we have the equation 300km = 1.5hr, we can just divide 1.5 (to isolate the number of hours) on both sides.

You should get 200 km = 1 hour

2.) Convert km to miles
To do this, you must know the conversion 1 km = 0.621371192 miles

So to convert 200 km to miles, just multiply 200 x 0.621371192.

You should get 124.274238.

That means the Ferrari drove at a pace of roughly 124 mph :)

8 0

3 years ago

Read 2 more answers

*High School problem not Middle School

devlian [24]

Should be c ? can you send me a picture please ? and it’s dice .

4 0

4 years ago

PLEASEE HELP MEEE I need help with my math homework