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stepladder [879]
2 years ago
12

Help please if very much appreciate it. ?

Mathematics
2 answers:
Alex777 [14]2 years ago
8 0

Answer:

7. 0.6

8. 0.45

9. 0.48

10. 0.28

11. 0.15

12. 0.75

LUCKY_DIMON [66]2 years ago
8 0
Same answer as the first one above !
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You rent an apartment that costs $900 per month during the first year, but the rent is
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Answer:1493.1

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2 years ago
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Can someone tell me what “increase s by 2s” is
Kryger [21]

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It means goes up by twos for example 2,4,6,8 all those numbers increase by 2s.

Step-by-step explanation:

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The base is 3 and the height is 2 what is the area of the triangle
Scorpion4ik [409]

Answer:

area = 3

Step-by-step explanation:

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6 0
2 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
Use the discriminant to find the number and type of solution for x^2+6x=-9​
likoan [24]

Answer:

D = 0; one real root

Step-by-step explanation:

Discriminant Formula:

\displaystyle \large{D =  {b}^{2}  - 4ac}

First, arrange the expression or equation in ax^2+bx+c = 0.

\displaystyle \large{ {x}^{2}  + 6x =  - 9}

Add both sides by 9.

\displaystyle \large{ {x}^{2}  + 6x + 9 =  - 9 + 9} \\  \displaystyle \large{ {x}^{2}  + 6x + 9 =  0}

Compare the coefficients so we can substitute in the formula.

\displaystyle \large{a {x}^{2}  + bx + c =  {x}^{2}  + 6x + 9 }

  • a = 1
  • b = 6
  • c = 9

Substitute a = 1, b = 6 and c = 9 in the formula.

\displaystyle \large{D =  {6}^{2}  - 4(1)(9)} \\  \displaystyle \large{D =  36 - 36} \\  \displaystyle \large{D =  0}

Since D = 0, the type of solution is one real root.

6 0
2 years ago
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