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4 years ago

Approximately 4/5 ofd US home owners have cell phones. What percent of homeowners do not have cell phones?

Mathematics

2 answers:

zheka24 [161]4 years ago

5 0

If 4/5 do, then 1/5 dont
1/5=x/100
x=20
20%

telo118 [61]4 years ago

5 0

It would have to be 1/5 because ofd is 4/5

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What is the annual interest rate if $1600 is invested for 6 years and $456 in interest is earned

sergij07 [2.7K]

Use the formula i = p*r*t:
$456
$456 = $1600*r*6. Solving for r, r = ----------------- = 0.0475, or 4 3/4%
($1600)(6)

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3 years ago

In a certain region, the equation y=19.485x+86.912 models the amount of a homeowner’s water bill, in dollars, where x is the n

DanielleElmas [232]

Answer:

It's choice 2.

Step-by-step explanation:

y=19.485x+86.912

The 19.485 is the slope of the graph of this equation. This gives the rate of change of the amount of the bill (above $86.912) for each added resident (x).

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3 years ago

Convert 2 3/4 into a improper faction

STatiana [176]

Answer:

11/4

Step-by-step explanation:

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3 years ago

Read 2 more answers

Please help w this! Its a calculus question! look at the picture for the problem,

neonofarm [45]

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 $x=\pm3\sqrt2$ and $y=\pm3\sqrt2$ .

Let $R$ be the region bounded by the line $x=3\sqrt2$ and the circle $x^2+y^2=36$ (the rightmost blue region). The right side of the circle can be expressed in terms of $x$ as a function of $y$ :

$x^2+y^2=36\implies x=\sqrt{36-y^2}$

Then the area of this circular segment is

$\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy$

Substitute $y=6\sin t$ , so that $\mathrm dy=6\cos t\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18$

Then the area of the entire blue region is 4 times this, a total of $\boxed{36\pi-72}$ .

Alternatively, you can compute the area of $R$ in polar coordinates. The line $x=3\sqrt2$ becomes $r=3\sqrt2\sec\theta$ , while the circle is given by $r=6$ . The two curves intersect at $\theta=\pm\dfrac\pi4$ , so that

$\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18$

so again the total area would be $36\pi-72$ .

Or you can omit using calculus altogether and rely on some basic geometric facts. The region $R$ is a circular segment subtended by a central angle of $\dfrac\pi2$ radians. Then its area is