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3 years ago

Hi pls help me!! It's due in 5 min

Mathematics

1 answer:

statuscvo [17]3 years ago

6 0

Answer:

Step-by-step explanation:

(if u need to show work) divide all options by 15 and 18 and whatever has a whole number the quickest will be the answer in this case 90

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This table shows values represented by an exponential function.

Bad White [126]

Answer:

$a = 2 , f(2)=9$

$b = 4 , f(4)=64$

And replacing the data into the average rate formula we got:

$m = \frac{64-9}{4-2}= 27.5$

And then the best answer for this case would be:

C. 27.5

Step-by-step explanation:

For this cae we know that the average rate of change of a function is given by this general expresion:

$m = \frac{f(b) -f(a)}{b-a}$

For this special case from the info of the table we have:

$a = 2 , f(2)=9$

$b = 4 , f(4)=64$

And replacing the data into the average rate formula we got:

$m = \frac{64-9}{4-2}= 27.5$

And then the best answer for this case would be:

C. 27.5

7 0

4 years ago

What strategies do you use when solving for an unknown number

frez [133]

Answer:

adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.

Step-by-step explanation:

please mark brainleist

5 0

2 years ago

If f(x) = 1 - 2x^3, find the inverse of f.

Artyom0805 [142]

Swap x and y's positions

f(x) is another way of saying y, so basically tha initial function is y = 1-2x^3.

Change it up to x = 1-2y^3 and solve for y

X - 1 = - 2y^3

(X - 1)/-2 = y^3

So the cube root of ((x-1)/2) = y

7 0

3 years ago

Determine whether the ffunction X/(x^2-1) is odd even or neither

djyliett [7]

equation is:

$f(x) = \frac{x}{ {x}^{2} - 1 }$

so if its even f(-x) should be equal to f(x)

that means

$f( - x) = \frac{ - x}{ {( - x)}^{2} - 1} \\ \\ f( - x) = - \frac{ x}{ {x}^{2} - 1}$

so f(x) is not equal to f(-x) so its not even

If its odd than f(-x) should be equal to -f(x)

$- f(x) = - ( \frac{ x}{ { x}^{2} - 1 } ) \\ - f(x) = - \frac{ x}{ {x}^{2} - 1 }$

so as we can see f(-x)= - f(x)

so the function is odd

Ans:function is odd and not even

3 0

3 years ago

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0

4 years ago