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3 years ago

Find the 20th term from the last term of the AP:3,8,13,..., 253.

Mathematics

1 answer:

RoseWind [281]3 years ago

3 0

Answer:

158

Step-by-step explanation:

The sequence is 3, 8, 13, ..., 253.

Going backwards, it's 253, 248, 243, ..., 3.

First term is 253, common difference is -5.

The nth term is:

a = 253 − 5(n − 1)

The 20th term is:

a = 253 − 5(20 − 1)

a = 158

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Step-by-step explanation:

Because 3+3 is 6 and 4+4 is 8

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3 years ago

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Elanso [62]

Answer:

87.8 square meters

Step-by-step explanation:

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To find the area of the two triangles, you could do 3*2.6. Normally, you would halve the values because they are triangles but since there's two of them it cancels out.

3*2.6 is 7.8, and adding that to the area of the rectangles gets the equation 81+7.8 which comes out to 87.8 square meters.

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3 years ago

Find an equation of the circle that satisfies the given conditions. (Give your answer in terms of x and y.)

masya89 [10]

Answer:

$(x-4)^2+(y+5)^2=90$

Step-by-step explanation:

The equation of a circle of radius r, centered at the point (a,b) is

$(x-a)^2+(y-b)^2=r^2$

We already know the center is at $(4,-5)$ , we are just missing the radius. To find the radius, we can use the fact that the circle passes through the point (7,4), and so the radius is just the distance from the center to this point (see attached image). So we find the distance by using distance formula between the points (7,4) and (4,-5):

radius $=\sqrt{(7-4)^2+(4-(-5))^2}=\sqrt{3^2+9^2}=\sqrt{90}$

And now that we know the radius, we can write the equation of the circle:

$(x-4)^2+(y-(-5))^2=\sqrt{90}^2$

$(x-4)^2+(y+5)^2=90$

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3 years ago

Solve the equation 3x=5(x+3)-3

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3 years ago

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

$\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18$

so the total area is, once again, $36\pi-72$ .

An even simpler way is to subtract the area of the square from the area of the circle.

$\pi6^2-(6\sqrt2)^2=36\pi-72$

6 0

3 years ago

Find the 20th term from the last term of the AP:3,8,13,..., 253. ​

Find the 20th term from the last term of the AP:3,8,13,..., 253.