Can I have help please

Please answer the equation 5n+1/8=1/2

Mamont248 [21]

$5n^{(8}+\frac{1}{8}=\frac{1}{2}^{(4} \\\\ 40n+1=4 \\\\ 40n=4-1 \\\\ 40n=3\\\\ \boxed{n=\frac{3}{40}}$

4 0

4 years ago

Read 2 more answers

Brainliest is given to the correct answer!

OlgaM077 [116]

Answer:

a. $p = 2\pi r + 8r$

b. $r = \frac{p}{2\pi + 8}$

Step-by-step explanation:

a. Radius of semicircle (r) = r

Diameter of semicircle (d) = 2r

Length of rectangle (l) = 2*diameter of semicircle = 2*2r = 4r

Distance around the track (p) = circumference of circle + 2(l)

Note: the two semicircles of the track = 1 full circle

Circumference of full circle = πd = π*2r = 2πr

Distance around the track:

p = 2πr + 2(4r)

p = 2πr + 8r

b. Rewriting the formula to make radius, r, the subject of the formula in terms of distance around the track.

$p = 2\pi r + 8r$

Factor out r

$p = r(2\pi + 8)$

Divide both sides by (2π + 8)

$\frac{p}{2\pi + 8} = \frac{r(2\pi + 8)}{2\pi + 8}$

$\frac{p}{2\pi + 8} = r$

$r = \frac{p}{2\pi + 8}$

6 0

4 years ago

A stadium can hold 20,000 people when it is full. The table below shows the

Kisachek [45]

If full then the stadium could have held 60000 people over 3 days.
It really held 17563+18126+16618 over the 3 days = 52307 people. To work out the difference 60000 - 52307 = 7693
Therefore 7693 seats were free over the 3 day period and so 7693 more people could have attended the concerts overall

3 0

4 years ago

A regular hexagonal pyramid with a base area of 166.3 square centimeters, a base side length of 8 centimeters, and a slant heigh

Nostrana [21]

I dont think you finished typing? or should the answer be obvious.... it is not obvious to me. Can i hav3 a hint

5 0

3 years ago

What are the limits of integration if the summation the limit as n goes to infinity of the summation from k equals 1 to n of the

Fofino [41]

Answer:

$\int_{2}^{9}x^2 dx$ so the limits are 2 and 9

Step-by-step explanation:

We want to express $\lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2$ as a integral. To do this, we have to identify $\sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2$ as a Riemann Sum that approximates the integral. (taking the limit makes the approximation equal to the value of the integral)

In general, to find a Riemann sum that approximates the integral of a function f over an interval [a,b] we can the interval in n subintervals of equal length and approximate the area (integral) with rectangles in each subinterval and them sum the areas. This is equal to

$\sum_{k=1}^n f(y_k) \frac{b-a}{n}$ , where $y_k\in[a+(k-1)\frac{b-a}{n},a+k\frac{b-a}{n}]$ is a selected point of the subinterval.

In particular, if we select the ending point of each subinterval as the $y_k$ , the Riemann sum is:

$\sum_{k=1}^n f(a+k\frac{b-a}{n}) \frac{b-a}{n}$ .

Now, let's identify this in $\sum_{k=1}^n\frac{1}{7n}(2+\frac{7k}{n})^2$ .

The integrand is x² so this is our function f. When k=n, the summand should be $\frac{b-a}{n}f(b)=\frac{b-a}{n}b^2$ because the last selected point is b. The last summand is $\frac{7}{n}(9)^2$ thus b=9 and b-a=7, then 9-a=7 which implies that a=2.

To verify our answer, note that if we substitute a=2, b=9 and f(x)=x² in the general Riemann Sum, we obtain the sum inside the limit as required.

4 0

3 years ago