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3 years ago

How to divide 48 by 5/8

1 answer:

4 0

Use the KCF method, or keep, change, flip.

1 ) Set up your expression: 48 / 5/8
2 ) Change the division symbol into multiplication : 48 x 5/8
3 ) Flip 5/8 : 48 x 8/5
4 ) Solve! If it helps,write 48 as 48/1. Just multiply as you would.

Answer: 384/5, or 76 4/5

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Nikki has $1100 and spends $35 per week. Mich has $520 and saves $55 per week. In about how many weeks, will Nikki and Micah hav

Rashid [163]

Answer:

6 weeks

Step-by-step explanation:

Given that :

Nikki :

Initial amount = $1100 ;

Amount spent per week = $35

Mich:

Initial amount = $520

Amount saved per week = $55

Number of weeks they'll both have the same amount :

Let number of weeks = w

For Nikki who is spending :

1100 - 35w - - - (1)

For Mich who is saving :

520 + 55w - - - (2)

Equate (1) and (2)

1100 - 35w = 520 + 55w

1100 - 520 = 55w + 35w

580 = 90w

w = 580 / 90

w = 6.444

About 6 weeks

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2 years ago

Find the value of x. Then find the measures of four angles.

posledela

Refer to the attached file

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3 years ago

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lozanna [386]

Answer:

EQUAL TO

Step-by-step explanation:

3 0

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