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2 years ago

How would the solution of the inequality 3/7 (35x - 14) ≤ 21x/2 + 3 be graphed on a number line?

Mathematics

1 answer:

Mashcka [7]2 years ago

4 0

Answer it's a closed dot on positive 2 going to the left

Step-by-step explanation:

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Find all possible solutions for x and y such that △ABC is congruent to △DEF.

Oxana [17]

Answer:

hello, this is you math teacher! i am undercover on this app and you have been banned for cheating

Step-by-step explanation:

jkjk lol

3 0

2 years ago

Read 2 more answers

) 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

Here are the results of a survey of 525 people in a town.

Ipatiy [6.2K]

525-238= 287 people speak Welsh
287-135 = 152 males speak welsh

238-163 = 75 females do not speak welsh
135+75 = 210 total females
525-210= 315 total males

152/315= 0.48 ~~ 48%

4 0

1 year ago

Select the correct answer.

tensa zangetsu [6.8K]

Answer:

7^1/3 * y^1/3

Step-by-step explanation:

(7y) ^ 1/3

We know that (ab) ^c = a^c * b^c

7^1/3 * y^1/3

7 0

3 years ago

Read 2 more answers

Make g the subject of this relation

zhannawk [14.2K]

Answer:

see explanation

Step-by-step explanation:

Simplify the radical

$\frac{U}{\frac{1}{f}+\frac{1}{g} }$

= $\frac{U}{\frac{g+f}{fg} }$

= $\frac{Ufg}{g+f}$

Square both sides

T² = $\frac{Ufg}{g+f}$ ( multiply both sides by (g + f) )

T²(g + f) = Ufg ( distribute left side )

T²g + T²f = Ufg ← subtract Ufg from both sides

T²g - Ufg + T²f = 0 ← subtract T²f from both sides

T²g - Ufg = - T²f ← factor out g from each term on the left side

g(T² - Uf) = - T²f ← divide both sides by (T² - Uf)

g = - $\frac{T^2f}{T^2-Uf}$ = $\frac{T^2f}{Uf-T^2}$

6 0

3 years ago