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sveta [45]
3 years ago
11

The rectangle shown has a perimeter

Mathematics
1 answer:
tresset_1 [31]3 years ago
8 0

L = 5+ 2W

LW= 250

you really only need 2 equations

but if you want you could also put

L+W = 35 cm

W= 10

L= 25

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Please help! Will mark the brainliest!
sesenic [268]
The domain is the set of all possible x-values which will make the function valid.
f(x) = \frac{3}{x-2} \ \ \ \ , \ g(x) =  \sqrt{x-1}
For the given function :
The denominator of a fraction cannot be zero
The number under a square root sign must be Non negative


(a)

(1) The domain of f ⇒⇒⇒ R - {2}

Because ⇒⇒⇒  x - 2 = 0  ⇒⇒⇒ x = 2

(2) The domain of g ⇒⇒⇒ [1,∞)
Because: x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1

(3) f + g = \frac{3}{x-2} + \sqrt{x-1}
The domain of (f+g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1


(4) f - g = \frac{3}{x-2} - \sqrt{x-1}
The domain of (f-g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1


(5) f * g = \frac{3}{x-2} * \sqrt{x-1} = \frac{3 \sqrt{x-1}}{(x-2)}
The domain of (f*g) ⇒⇒⇒ [1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 ≥ 0 ⇒⇒⇒ x ≥ 1

(6) f * f = \frac{3}{x-2} * \frac{3}{x-2} = \frac{9}{(x-2)^2}
The domain of ff ⇒⇒⇒ R - {2}

Because ⇒⇒⇒  x-2 = 0  ⇒⇒⇒ x = 2

(7) \frac{f}{g} =   \frac{\frac{3}{x-2} }{ \sqrt{x-1} } =  \frac{3}{(x-2) \sqrt{x-1}}
The domain of (f/g) ⇒⇒⇒ (1,∞) - {2}
because: x-2 = 0 ⇒⇒⇒ x = 2   and    x - 1 > 0 ⇒⇒⇒ x > 1

(8) \frac{g}{f} =  \frac{ \sqrt{x-1} }{ \frac{3}{x-2} } =  \frac{1}{3} (x-2) \sqrt{x-1}
The domain of (g/f) ⇒⇒⇒ [1,∞) - {2}
Because: x - 2 = 0 ⇒⇒⇒ x = 2   and   x - 1 ≥ 0  ⇒⇒⇒ x ≥ 1
===================================================
(b)


(9) (f + g)(x) = \frac{3}{x-2} + \sqrt{x-1}


(10) (f - g)(x) = \frac{3}{x-2} - \sqrt{x-1}


(11) (f * g)(x) = \frac{3}{x-2} * \sqrt{x-1} = \frac{3 \sqrt{x-1}}{(x-2)}


(12) (f * f)(x) = \frac{3}{x-2} * \frac{3}{x-2} = \frac{9}{(x-2)^2}


(13) \frac{f}{g} =   \frac{\frac{3}{x-2} }{ \sqrt{x-1} } =  \frac{3}{(x-2) \sqrt{x-1}}


(14) (\frac{g}{f})(x) =  \frac{ \sqrt{x-1} }{ \frac{3}{x-2} } =  \frac{1}{3} (x-2) \sqrt{x-1}



7 0
3 years ago
Solve the absolute value equation. |3x| = 12 A. {4, 4} B. {4, –4} C. {–4, –4} D. no solution
AlekseyPX
<span>|3x| = 12 actually represents two equations:  3x = 12 and -3x = 12.

Thus, the solutions are {-4, 4).</span>
8 0
3 years ago
Easy Question pls help. (Algebra II type problems)
Roman55 [17]

Answer:

\large\boxed{if\ x\neq-3\ \wedge\ y\neq0\ \wedge\ x\neq y\ \wedge\ x\neq -y}

second

\large\boxed{y\neq0\ \wedge\ x\neq-y\ (y\neq-x)}

Step-by-step explanation:

We know that dividing by 0 is impossible.

Therefore, the denominator of the expression must be different from 0.

\text{For}\\\\\dfrac{(x-y)^2}{2xy+6y}\times\dfrac{4x+12}{x^2-y^2}\\\\\\2xy+6y\neq0\qquad(1)\\\\x^2-y^2\neq0\qquad(2)

(1)\\2xy+6y\neq0\qquad\text{distribute}\\\\2y(x+3)\neq0\iff2y\neq0\ \wedge\ x+3\neq0\\\\2y\neq0\qquad\text{divide both sides by 2}\\\boxed{y\neq0}\\\\x+3\neq0\qquad\text{subtract 3 from both sides}\\\boxed{x\neq-3}\\==========================\\(2)\\x^2-y^2\neq0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\(x-y)(x+y)\neq0\iff x-y\neq0\ \wedge\ x+y\neq0\\\boxed{x\neq y}\ \wedge\ \boxed{x\neq-y}

\text{For}\\\\\dfrac{2(x-y)}{y(x+y)}\\\\y(x+y)\neq0\iff y\neq0 \wedge\ x+y\neq0\\\\ \boxed{y\neq0}\ \wedge\ \boxed{x\neq-y}

3 0
3 years ago
Describe the relationship between the values of the digits in the number 9,999,999
Anit [1.1K]
Because all the digits are the same, each digit is worth 10x the one to its right and 1/10 of the one to its left.
This is because we count in base 10. This is also why we can expand this number to become
9*10^6+ 9*10^5+ 9*10^4+ 9*10^3+ 9*10^2+ 9*10^1 +9*10^0.
6 0
3 years ago
Read 2 more answers
Dion practiced the long jump . He jumped 3.05 meters,2.74 meters and 3.3 meters.What was the average length of Dion's jumps?
Mademuasel [1]
All you do is subtract 3.05 and 2.74 then do the same thing with the other ones and see what number is between all of them. What grade are you in
5 0
3 years ago
Read 2 more answers
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