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2 years ago

Solve the inequality- x+18 ≥ 8x+4 or 15x-15 ≤ 15x+5 I give Brainliest!

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

6 0

Answer:

x ≥ 2

0 ≤ 20 (not quite sure if the question is right)

Step-by-step explanation:

x+18 ≥ 8x+4

x - 8x ≥ 4 - 18

-7x ≥ -14

x ≥ $\frac{-14}{-7}$

x ≥ 2

________________

15x-15 ≤ 15x+5

15x - 15x ≤ 5 + 15

0 ≤ 20

(I don't think this is the right question...cuz theres a 0)

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Find the formula for the area of the shaded region in the figure.<br> x² = 4py<br> ہے

Alexandra [31]

The formula for the area of the shaded region in the figure.

x² = 4py is mathematically given as

$A=\frac{8}{3} \sqrt{p h} \cdot h$

<h3>What is the formula for the area of the shaded region in the figure?</h3>

Parameters

$x^{2}=4 p y$

$y=h$

$\therefore x^{2}=4 p h \Rightarrow x=2 \sqrt{p h} \\$

Generally, the equation for the Area is mathematically given as

$& A=2 \int_{0}^{2 \sqrt{p h}}\left(h-\frac{x^{2}}{4 p}\right) d x \\\\& A=2\left(h x-\frac{x^{3}}{12 p}\right]_{0}^{2 \sqrt{p h}} \\\\& A=2\left[\left[h 2 \sqrt{p h}-\frac{1}{12 p} \cdot(2 \sqrt{p h})^{3}-0\right]\right. \\$

$&A=2\left[2 h^{3 / 2} \sqrt{p}-\frac{8(\sqrt{p})^{3}(\sqrt{h})^{3}}{12 p}-0\right] \\\\&A=2\left[2 h^{k / 2} \sqrt{p}-\frac{2}{3} \sqrt{p} \cdot h^{3 / 2}\right] \\\\&A=2 \times \frac{4 \sqrt{p} \cdot h^{3 / 2}}{3} \\\\&A=\frac{8}{3} \cdot \sqrt{p} \cdot \sqrt{h} \cdot h \\\\&A=\frac{8}{3} \sqrt{p h} \cdot h$

$A=\frac{8}{3} \sqrt{p h} \cdot h$

In conclusion, the equation for the Area is

$A=\frac{8}{3} \sqrt{p h} \cdot h$

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