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levacccp [35]
3 years ago
5

2.2 * 10^-2 * 4 * 10^4 i need help asap im really confused

Mathematics
1 answer:
Westkost [7]3 years ago
8 0

If my math is correct and I understood the question correctly then the answer is 880.

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Please help. I don't understand it. HELP ASAP....
Vladimir [108]

Answer:

\large\boxed{x\leq-27}

Step-by-step explanation:

\dfrac{x}{-9}\geq3\qquad\text{change the signs}\\\\\dfrac{x}{9}\leq-3\qquad\text{multiply both sides by 9}\\\\9\!\!\!\!\diagup^1\cdot\dfrac{x}{9\!\!\!\!\diagup_1}\leq(-3)(9)\\\\x\leq-27

5 0
3 years ago
Solve the following equation for all values of x <br> 2x^2+18x-17=11x-2
Cerrena [4.2K]

2x^2 + 18x - 17 = 11x - 2

           -11x   +2    -11x   +2

2x^2 +7x - 15 = 0

(x - 1.5) (x + 5)

x = 1.5    x = -5

3 0
3 years ago
What is the radius of a circle with an area of 530.9
Igoryamba

Answer:

13

Step-by-step explanation:

diameter = r * 2 = 13  * 2 = 26 (diameter is double radius)

Circle area = π * r^2 = π * 169 ≈ 530.9

3 0
3 years ago
Read 2 more answers
What's greater 3 fourths or 2 fourths
Len [333]

3  of anything positive is greater than  2  of the same thing. 
I can't think of an exception.

7 0
3 years ago
Read 2 more answers
Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
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