Answer:
8 inches
Step-by-step explanation:
Let w represent the width.
If the length is 14 inches longer than the width, it can be represented by w + 14.
Use the perimeter formula, p = 2l + 2w. Plug in 60 as the perimeter and w + 14 as l, then solve for w:
p = 2l + 2w
60 = 2(w + 14) + 2w
60 = 2w + 28 + 2w
60 = 4w + 28
32 = 4w
8 = w
So, the width is 8 inches
Answer:
f(x) =
+ 1
Step-by-step explanation:
Whenever addition or subtraction occurs within the cube root, it moves horizontally; the opposite is done for the translation, thus if it is addition, which in this case it is, instead of moving the positive direction, the function moves the negative direction.
As opposed to addition or subtraction outside of the cube root, it moves vertically, and in this case, it is translated exactly as stated. Here it states addition, thus the graph moves up in one unit.
The origin of the parent function is at ( 0, 0), in contrast, the origin of the function is at ( -6, 1 )
By using what we know, we can determine that the answer will be f(x) =
+ 1 since we know that the inside of the cube root should be the opposite of the x value and the outside of the cube root should be the same.
Since you mentioned calculus, perhaps you're supposed to find the area by integration.
The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines
and
.
Let
be the region bounded by the line
and the circle
(the rightmost blue region). The right side of the circle can be expressed in terms of
as a function of
:
![x^2+y^2=36\implies x=\sqrt{36-y^2}](https://tex.z-dn.net/?f=x%5E2%2By%5E2%3D36%5Cimplies%20x%3D%5Csqrt%7B36-y%5E2%7D)
Then the area of this circular segment is
![\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_R%5Cmathrm%20dA%3D%5Cint_%7B-3%5Csqrt2%7D%5E%7B3%5Csqrt2%7D%5Cint_%7B3%5Csqrt2%7D%5E%7B%5Csqrt%7B36-y%5E2%7D%7D%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy)
![=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_%7B-3%5Csqrt2%7D%5E%7B3%5Csqrt2%7D%28%5Csqrt%7B36-y%5E2%7D-3%5Csqrt2%29%5C%2C%5Cmathrm%20dy)
Substitute
, so that ![\mathrm dy=6\cos t\,\mathrm dt](https://tex.z-dn.net/?f=%5Cmathrm%20dy%3D6%5Ccos%20t%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_%7B-%5Cpi%2F4%7D%5E%7B%5Cpi%2F4%7D6%5Ccos%20t%28%5Csqrt%7B36-%286%5Csin%20t%29%5E2%7D-3%5Csqrt2%29%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_%7B-%5Cpi%2F4%7D%5E%7B%5Cpi%2F4%7D%2836%5Ccos%5E2t-18%5Csqrt2%5Ccos%20t%29%5C%2C%5Cmathrm%20dt%3D9%5Cpi-18)
Then the area of the entire blue region is 4 times this, a total of
.
Alternatively, you can compute the area of
in polar coordinates. The line
becomes
, while the circle is given by
. The two curves intersect at
, so that
![\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_R%5Cmathrm%20dA%3D%5Cint_%7B-%5Cpi%2F4%7D%5E%7B%5Cpi%2F4%7D%5Cint_%7B3%5Csqrt2%5Csec%5Ctheta%7D%5E6r%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
![=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cfrac12%5Cint_%7B-%5Cpi%2F4%7D%5E%7B%5Cpi%2F4%7D%2836-18%5Csec%5E2%5Ctheta%29%5C%2C%5Cmathrm%20d%5Ctheta%3D9%5Cpi-18)
so again the total area would be
.
Or you can omit using calculus altogether and rely on some basic geometric facts. The region
is a circular segment subtended by a central angle of
radians. Then its area is
![\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18](https://tex.z-dn.net/?f=%5Cdfrac%7B6%5E2%5Cleft%28%5Cfrac%5Cpi2-%5Csin%5Cfrac%5Cpi2%5Cright%29%7D2%3D9%5Cpi-18)
so the total area is, once again,
.
An even simpler way is to subtract the area of the square from the area of the circle.
![\pi6^2-(6\sqrt2)^2=36\pi-72](https://tex.z-dn.net/?f=%5Cpi6%5E2-%286%5Csqrt2%29%5E2%3D36%5Cpi-72)
The variable that cannot be illustrated by a stem-and-leaf plot is C. Number of doctor visits during the past year.
<h3>What is a stem and leaf plot?</h3>
It can be inferred that a stem and leaf plot simply means the technique that is used to classify continuous or discrete variables.
In this case, the variable that cannot be illustrated by a stem-and-leaf plot is the number of doctor visits during the past year. This is because it's an ordinal variable.
Learn more about stem and leaf plot on:
brainly.com/question/12276901
Answer:
I would love to help you so much but I don't understand