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3 years ago

Find the value of the variable and GH if H is between G and I. GI=5b+2,HI=4b-5, HI=3

Mathematics

1 answer:

gulaghasi [49]3 years ago

4 0

Answer:

GH = 9 units

Step-by-step explanation:

Given HI = 4b - 5 and HI 3, then

4b - 5 = 3 ( add 5 to both sides )

4b = 8 ( divide both sides by 4 )

b = 2

Thus

GI = 5b + 2 = 5(2) + 2 = 10 + 2 = 12

GH = GI - HI = 12 - 3 = 9

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) 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.

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3 years ago

1. Jane went shopping and bought a shirt for $27, a skirt for $32, and a pair of shorts for $14. Choose the answer that is the b

Natasha2012 [34]

Answer:

Step-by-step explanation:

First add the tens: 20 + 30 + 10= 60

Now add the Ones: 7 + 2 + 4= 13

now add them all together: 60+13=73

so the best estimate is 70

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3 years ago

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andrey2020 [161]

Answer:

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Step-by-step explanation:

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3 years ago

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Anarel [89]

Answer:

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3 years ago

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Find the value of the variable and GH if H is between G and I. GI=5b+2,HI=4b-5, HI=3​

Find the value of the variable and GH if H is between G and I. GI=5b+2,HI=4b-5, HI=3