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den301095 [7]
3 years ago
5

Om to 4conm.

Mathematics
1 answer:
max2010maxim [7]3 years ago
8 0
The correct answer is water
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What is one way to find the number of grams in 6 kilograms?​
gavmur [86]

Answer:

Multiply by 1000

Step-by-step explanation:

6kg equals 6000g

8 0
3 years ago
What is the simplified form of 3.1 - 5.8n - 4n + 8?
lana66690 [7]

Answer:

-9.8n + 11.1

Step-by-step explanation:

1. Combine like terms

      -5.8n - 4n = -9.8n

      3.1 + 8 = 11.1

2. -9.8n +11.1

8 0
2 years ago
Read 2 more answers
Solve 3x-1 =2x+1 solve the question ​
galina1969 [7]

3x-1=2x+1

3x-2x = 1+1

x=2

8 0
2 years ago
Read 2 more answers
When solving p^2 + 5 = 8p - 7. Kate wrote p^2 + 12 = 8p. The property she used was:
spayn [35]

Answer:

4)

Step-by-step explanation:

Because they just added the same equality. And when you move the -7 over the equal sign it makes it positive so it would be p^2 + 5 + 7 = 8p and that makes it p^2 + 12 = 8p

6 0
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
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