3 years ago

The perimeter of a rectangle is 24 feet. The length is 3 feet less than twice the width. The width is:

1 answer:

8 0

the answer is B cause 5 times 2 is 10 and 10-3=7

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LuckyWell [14K]

180(n-2)=n*144 in our casesolving for n we obtain:36n = 360n=10 sidesCHECK!!!!!!!!!!!!!!!!!!!

4 0

3 years ago

Dima020 [189]

It would be 6x^2 + 14x -12

8 0

2 years ago

AleksAgata [21]

I think u can just divide by 16 to get it because each ounce is 16

8 0

3 years ago

gavmur [86]

Answer:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{x^2+y^2+49}+7\right)=14$

Step-by-step explanation:

to find the limit:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{x^2+y^2}{\sqrt{x^2+y^2+49}-7}\right)$

we need to first rationalize our expression.

$\dfrac{x^2+y^2}{\sqrt{x^2+y^2+49}-7}\left(\dfrac{\sqrt{x^2+y^2+49}+7}{\sqrt{x^2+y^2+49}+7}\right)$

$\dfrac{(x^2+y^2)(\sqrt{x^2+y^2+49}+7)}{(\sqrt{x^2+y^2+49}\,)^2-7^2}$

$\dfrac{(x^2+y^2)(\sqrt{x^2+y^2+49}+7)}{(x^2+y^2)}$

$\sqrt{x^2+y^2+49}+7$

Now this is our simplified expression, we can use our limit now.

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{x^2+y^2+49}+7\right)\\\sqrt{0^2+0^2+49+7}\\7+7\\14$

Limit exists and it is 14 at (0,0)

4 0

3 years ago

choli [55]

Answer:

it is b.5/36

Step-by-step explanation:

5 0

3 years ago