Find the value of f(-8).

Apply the square root principle to solve (x - 2)2 + 20 = 0.

sashaice [31]

OD) x =-2 + 215, = -2 - 2 15

8 0

3 years ago

Can you guys help me? Please and thank you! :)

expeople1 [14]

The theoretical probability is what is expected based on using math. Since there is only one card that is number 1, and there are 400 possible outcomes, then the theoretical probability is 1/400 which is 0.25%.
Now find the experimental probability by find the number of times card 1 is picked, and the number of cards in total. 128/400 = 0.32 which is 32 percent.
The experimental probability is higher than the theoretical probability.
Hope this helps

4 0

2 years ago

Zack buys some oranges from a store. The oranges cost $4.00 for 2 pounds. He wants to buy more fruit, but at the same price per

SSSSS [86.1K]

Answer:

C

Step-by-step explanation:

The oranges cost $4 per 2 lbs which is 4/2=$2 per lb. Option C has peaches at $6 for 3 lbs which is 6/3 =$2 per lb.

5 0

3 years ago

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For the function given below, find a formula for the Riemann sum obtained by dividing the interval [0,5] into n equal subinterva

sergij07 [2.7K]

Given

we are given a function

$f(x)=x^2+5$

over the interval [0,5].

Required

we need to find formula for Riemann sum and calculate area under the curve over [0,5].

Explanation

If we divide interval [a,b] into n equal intervals, then each subinterval has width

$\Delta x=\frac{b-a}{n}$

and the endpoints are given by

$a+k.\Delta x,\text{ for }0\leq k\leq n$

For k=0 and k=n, we get

$\begin{gathered} x_0=a+0(\frac{b-a}{n})=a \\ x_n=a+n(\frac{b-a}{n})=b \end{gathered}$

Each rectangle has width and height as

$\Delta x\text{ and }f(x_k)\text{ respectively.}$

we sum the areas of all rectangles then take the limit n tends to infinity to get area under the curve:

$Area=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)$

Here

$f(x)=x^2+5\text{ over the interval \lbrack0,5\rbrack}$ $\Delta x=\frac{5-0}{n}=\frac{5}{n}$ $x_k=0+k.\Delta x=\frac{5k}{n}$ $f(x_k)=f(\frac{5k}{n})=(\frac{5k}{n})^2+5=\frac{25k^2}{n^2}+5$

Now Area=

$\begin{gathered} \lim_{n\to\infty}\sum_{k\mathop{=}1}^n\Delta x.f(x_k)=\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{5}{n}(\frac{25k^2}{n^2}+5) \\ =\lim_{n\to\infty}\sum_{k\mathop{=}1}^n\frac{125k^2}{n^3}+\frac{25}{n} \\ =\lim_{n\to\infty}(\frac{125}{n^3}\sum_{k\mathop{=}1}^nk^2+\frac{25}{n}\sum_{k\mathop{=}1}^n1) \\ =\lim_{n\to\infty}(\frac{125}{n^3}.\frac{1}{6}n(n+1)(2n+1)+\frac{25}{n}n) \\ =\lim_{n\to\infty}(\frac{125(n+1)(2n+1)}{6n^2}+25) \\ =\lim_{n\to\infty}(\frac{125}{6}(1+\frac{1}{n})(2+\frac{1}{n})+25) \\ =\frac{125}{6}\times2+25=66.6 \end{gathered}$

So the required area is 66.6 sq units.

3 0

1 year ago

What can be factored from 12g²-27h²

Elenna [48]

Answer: You can use only common factor (3).

Step-by-step explanation:

12g²-27h²

3 . ( 4g² - 9h² )

4 0

3 years ago

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