All categories

3 years ago

The perimeter of a rectangular field is 284 m.

Mathematics

1 answer:

Fantom [35]3 years ago

7 0

Answer:

I think it would be 208

Step-by-step explanation:

Well if the length is 76 and the total perimeter is 284, I would think that all you would have to do is subtract 284 by 76 to get 208. Sorry if this is confusing never been good at explaining things.

Hope this helps! Have a wonderful rest of your evening! :)

~ Goldilocks

You might be interested in

-c: infinitely many solutions, Help solve plss

trapecia [35]

I think it’s b cos u minus 43 and your left with 48x and 47x so divide by 47 and you get 1 point something x = to x so yeah probably none because no number will make that work so it’s b

7 0

3 years ago

Find the missing side of this right

navik [9.2K]

Answer:

x = 13

Step-by-step explanation:

First start with the blank equation

$a^{2} + b^{2} = c^{2}$

Plug in the legs and hypotenuse. Remember c squared is always the hypotenuse.

$5^{2} + 12^{2} = c^{2}$

Then solve

25 + 144 = $c^{2}$

169 = $c^{2}$

$\sqrt{169} = \sqrt{c^{2} }$

c = 13

3 0

3 years ago

Find f(x)+g(x), given f(x)=x^2-8 and g(x)=-2-x^2

Gnoma [55]

$\left(x^2-8\right)+\left(-2-x^2\right)=-10$

$\left(x^2-8\right)+\left(-2-x^2\right)$

$=x^2-8-2-x^2$

$=x^2-x^2-8-2$

$=-8-2$

$=-10$

8 0

3 years ago

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0

3 years ago

What is the solution to the system of equations?

frez [133]

(-19,55) x is equal to -19 and y is equal to 55

4 0

3 years ago

Read 2 more answers