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3 years ago

9

(6a-3b^2-a) + (b-4+7a^2)Find the sum or difference

1 answer:

morpeh [17]3 years ago

3 0

6a-3b^2-a+b-4+7a^2
goup like terms
7a^2-3b^2+6a-a+b-4
7a^2-3b^+5a-4

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7x – 8y = 112 in slope intercept form

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Answer:

y = 7/8x-14

Step-by-step explanation:

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If cos theta=-3/5 in quadrant 2 what is sin theta?<br><br> A:4/5<br> B:3/4<br> C:-3/4<br> D:-4/5

Scrat [10]

Answer: Choice A) 4/5

-------------------------------------------

Work Shown:

cos^2(theta) + sin^2(theta) = 1
(-3/5)^2 + sin^2(theta) = 1
9/25 + sin^2(theta) = 1
9/25 + sin^2(theta) - 9/25 = 1 - 9/25
sin^2(theta) = 1 - 9/25
sin^2(theta) = 25/25 - 9/25
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3 years ago

) The density of oil in a circular oil slick on the surface of the ocean at a distance of r meters from the center of the slick

Feliz [49]

Answer:

Therefore the mass of the of the oil is 409.59 kg.

Step-by-step explanation:

Let us consider a circular disk. The inner radius of the disk be r and the outer diameter of the disk be (r+Δr).

The area of the disk

=The area of the outer circle - The area of the inner circle

= $\pi (r+\triangle r)^2- \pi r^2$

$=\pi [r^2+2r\triangle r+(\triangle r)^2-r^2]$

$=\pi [2r\triangle r+(\triangle r)^2]$

Since (Δr)² is very small, So it is ignorable.

∴ $A=2\pi r\triangle r$

The density $\delta (r)= \frac{40}{1+r^2}$

We know,

Mass= Area× density

$=(2r \pi \triangle r)(\frac{40}{1+r^2}})$

Total mass $M=\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i$

Therefore

$\sum_{i=1}^n \frac{80r_i\pi }{1+r^2}\triangle r_i=\int_0^5 \frac{80r\pi }{1+r^2}dr$

$=40\pi[ln(1+r^2)]_0^5$

$=40\pi [ln(1+5^2)-ln(1+0^2)]$

$=40\pi ln(26)$

= 409.59 kg (approx)

Therefore the mass of the of the oil is 409.59 kg.

3 0

2 years ago