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Veseljchak [2.6K]
4 years ago
13

Y=2x x=-y+6 What does it equal to

Mathematics
1 answer:
erik [133]4 years ago
4 0

Answer:

y=4

x=2

Step-by-step explanation:


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erik [133]
Top is 113cm squared
6 0
3 years ago
2 (-3x-4) &gt;16<br> Solve the inquality for X
n200080 [17]

Answer:

x < -4

Step-by-step explanation:

2(-3x - 4) > 16

-6x - 8 > 16

-6x > 16 + 8

-6x > 24

-x > 4

x < -4

3 0
3 years ago
Read 2 more answers
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
Ann takes 70 paces to walk 50m. the number of paces ann takes to walk 3.5km is A.70 B.490. C.3900 D.4900​
Shalnov [3]
<h3>Answer:  4900  (choice D)</h3>

===================================================

Work Shown:

1 km = 1000 m

3.5 km = 3500 m  ............. multiplying both sides by 3.5

(70 paces)/(50 m) = (x paces)/(3500 m)

70/50 = x/3500

7/5 = x/3500

7*3500 = 5x .............. cross multiply

5x = 7*3500

x = (7*3500)/5

x = (7*5*700)/5

x = 7*700

x = 4900

Ann takes 4900 paces to walk 3.5 km

4 0
2 years ago
Help please geometry !
ivanzaharov [21]
So here we can find out that the two angles are vertical angles. That means that they have the same measure. So what we need to do is set them equal to each other:

2x+20=3x-30
x=50

7 0
3 years ago
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