The questions is:Simplify and write the answers in exponential form.

Claire bought some books. One eighth of them are fiction. She has 2 fiction books. How many books does claire have in all.

kirza4 [7]

16 because one eighth is 2 and if u times 2 by 8 because it’s one eighth then u have 16

3 0

3 years ago

What is the sum of the arithmetic sequence 151, 137, 123,..if there are 26 terms?

ElenaW [278]

First term: a1 = 151
common difference: d = -14 (we decrease by 14 each time, eg, 151-14 = 137)

nth term of this arithmetic sequence is...
an = a1+d(n-1)
an = 151+(-14)(n-1)
an = 151-14n+14
an = -14n+165

This will be used in the formula below

Sn = n*(a1+an)/2
Sn = n*(151+(-14n+165))/2
S26 = 26*(151+(-14*26+165))/2 ... replace every n with 26
S26 = -624

The final answer here is choice C) -624

5 0

3 years ago

The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain riv

katrin2010 [14]

Answer:

7.25

Step-by-step explanation:

200m/s^6 = m/3^6

s^6 = 200*3^6

s = 3*200^(1/6) = 7.25 mi/hr

Hope I helped!

8 0

3 years ago

1st answer gets brainliest!!!

Nastasia [14]

The apothem of this is 13

7 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$