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kipiarov [429]
3 years ago
12

The area of a rectangle is 27m and the length of the rectangle is 3m less then twice the width. Find the dimensions of the recta

ngle
Mathematics
1 answer:
Zielflug [23.3K]3 years ago
5 0

Answer:

w = 4 1/2

l = 6

Step-by-step explanation:

w = width

2w-3 = length

A = lw

27 = w(2w-3)

27 = 2w² - 3w

subtract 27 from each side:

2w² - 3w - 27 = 0

factor:

(2w-9)(w+3) = 0

solve each for 'w':

w = 9/2

w = -3  (This answer can be discounted because a length cannot be negative)

width = 9/2

length = 2(9/2) - 3 which is 6

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DiKsa [7]

Firstly let's find hypotenuse(let it will be "n" of smaller triangle

Let use Pythagorean theorem

a^{2}  +  {b}^{2}  =  {n}^{2}  \\  {20}^{2}  +  {10}^{2}  =  {n}^{2}  \\  n =  \sqrt{500}

Now we need to find hypotenuse(x) of bigger triangle

{c}^{2}  +  {n}^{2}  =  {x}^{2}  \\  {9}^{2}  +  { \sqrt{500} }^{2}  =  {x}^{2}  \\ x =  \sqrt{581}

The value of x must be rounded to 1 DP, so

\sqrt{581}  = 24.1039... \\  \sqrt{581}  \simeq \: 24.1

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3 years ago
I forgot how to solve this can someone explain please
SIZIF [17.4K]

Answer:

The radius of cylinder is 14 inches

Step-by-step explanation:

Given that the volume of cylinder is 196π in² and the height is 1 in . The formula for it is V = πr²h. Then you can substitute the following value into the formula:

V = 196π

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196π = π × r² × 1

r² = 196π/π

r² = 196

r = 14 in

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3 years ago
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Mrs cousins works in a pool store and earns a 9.5% commission on every swimming pool she sells. how much commission did she earn
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Eduardwww [97]
Part A. You have the correct first and second derivative.

---------------------------------------------------------------------

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

-------------------------------------------------------------

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h  ' (x) = -10a(-2x+1)^4
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so either
-10a = 0 or (-2x+1)^4 = 0
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