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snow_tiger [21]
2 years ago
15

Let R = {0, 1, 2, 3} be the range of h (x ) = x - 7. The domain of h inverse is

Mathematics
1 answer:
denpristay [2]2 years ago
4 0
The range of a function is the domain of its inverse
so domain of inverse is {0,1,2,3}
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Triple a number, x, increased by 1 is between -20 and 10. What are the solutions for x?
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-20 \leq 3x+1 \leq 10\\
\Rightarrow\ -21 \leq 3x \leq 9\\
\Rightarrow\ -7 \leq x \leq 3\\

Sol=[-7\ 3]\ if\ x\ is\ a\ real.


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Is this tangent to the circle?
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What are the answers to these i put them in the picture
tekilochka [14]
The problem to number 1 is a inequality, so
x-2<-7
x<-5
since it is < it will have open dot.
Answer:  C

2.  x/3>-3
      x>-9

Answer: D

3.  5p+26<72
                5p<46
                 p<9.2

Answer: C

4.  -3w+3>=18
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5. 15x-21>= 12x+18
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8 0
3 years ago
Someone help ASAP!!!!!!
Alchen [17]

Answer:

Length, l = 11 ft.

Width, w = 9 ft.

Step-by-step explanation:

From the given data, the area of the rectangle = 99 ft².

Area of the rectangle = Length, l X Width, w

Here, Length, l = 7 more than twice the width

⇒       Length, l = 7 + 2w

Therefore, Area, A = 99 = (7 + 2w)w

⇒ 99 = 7w + 2w²

⇒ 2w² + 7w - 99 = 0

Solve the Quadratic equation using the formula: x = $ \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ for the quadratic equation ax² + bx + c = 0.

Therefore, w = $ \frac{-7 \pm \sqrt{7^2 -4(2)(-99)}}{2(2)} $

$ = \frac{-7 \pm \sqrt{49 + 8(99)}}{4} $

$ = \frac{-7 \pm + \sqrt{841}}{4} $

Since, $ \sqrt{841} = 29 $ we get:

$ w = \frac{-7 \pm 29}{4} $

This gives two values of 'w', viz., w = $ \frac{-7 - 29}{4} $, $ \frac{-7 + 29}{4} $

$ \implies w = \frac{22}{4}, \frac{-36}{4} $

⇒ w = $ \frac{22}{4} $, -9.

We take the integer values.

If w = -9, then l = 2(-9) + 7

⇒ l = - 18 + 7 = - 11

Therefore, the length, l of the rectangle = - 11 ft.

and the width, w of the rectangle = - 9 ft.

Hence, the answer.

3 0
2 years ago
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