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Furkat [3]
2 years ago
8

Can someone who understands this help me?????? Thank you so much!!!!!!!!

Mathematics
1 answer:
Assoli18 [71]2 years ago
8 0

what is f(-3) is another way to say "what is the value of f(x) when x = -3?", well for that we can simply look at the piece-wise function graphed, Check the picture below.

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An old rectangular tank with length 0.3 m could hold 154 L of water. A new similar tank has a length of 1.5 m. What is the capac
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The length of the new tank is 5 times bigger than the old one, so 154×5=770. As a result, the capacity of the new tank is 770L. Hope it help!
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Factor: 4x2 + 4x<br>Expand: -(x – 3)​
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<h2>Question: Factor: 4x2 + 4x</h2>

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gcf = 4x

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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}



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A pile of 15 boxes is 3 metres high. What is the depth of each box?
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