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3 years ago

Find the dimension of a rectangle with a perimeter of 56ft if L is 6ft greater than twice the W

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

3 0

Answer:

ita greater

Step-by-step explanation:

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3 years ago

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Solve the system of equations 2r - 6y= 28 -x-y = 14 PLEASE HELP

o-na [289]

Answer:

$\displaystyle [-7, -7]$

Step-by-step explanation:

{2x - 6y = 28

{−x - y = 14

−⅙[2x - 6y = 28]

{−⅓x + y = −4⅔

{−x - y = 14

____________

$\displaystyle \frac{-1\frac{1}{3}x}{-1\frac{1}{3}} = \frac{9\frac{1}{3}}{-1\frac{1}{3}} \\ \\$

$x = -7$ [Plug this back into both equations above to get the y-coordinate of −7]; $\displaystyle -7 = y$

I am joyous to assist you anytime.

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3 years ago

Three friends just out of culinary school have applied for a loan to open a new restaurant in the heart of New York City. The ba

blsea [12.9K]

Answer:

The answer is 0.3 million per year

Step-by-step explanation:

8 0

2 years ago

PLEASE HELP!!

erik [133]

$2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha$

$1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}$

$3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x$

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3 years ago

please help thank you

Free_Kalibri [48]

k(-3)=36(2)

hope it helps

7 0

2 years ago