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3 years ago

7 What is the solution set for - 4x -10 < 2?

Mathematics

1 answer:

alex41 [277]3 years ago

6 0

Answer:

$x\g{-3}$

Step-by-step explanation:

given $-4x-10\l{2}$

$-4x\l{12}$

$4x\g{-12}$

$x\g{-3}$

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4) A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is

lisov135 [29]

Answer:

l=0.1401P\\

w =0.2801P

where P = perimeter

Step-by-step explanation:

Given that a window is in the form of a rectangle surmounted by a semicircle.

Perimeter of window =2l+\pid/2+w

$P= 2l+3.14 (w/2)+w$

Or $P = 2l+2.57w\\l = \frac{P-2.57w}{2}$

To allow maximum light we must have maximum area

Area = area of rectangle + area of semi circle where rectangle width = diameter of semi circle

$A=lw +\pi \frac{w^2}{8}$

$A=lw +\pi \frac{w^2}{8}\\A=w*\frac{P-2.57w}{2}+0.3925w^2\\2A= Pw-2.57w^2+(0.785w^2)\\2A' = P-5.14w+1.57 w\\2A" =-5.14+1.57$

Hence we get maximum area when i derivative is 0

i.e. $P-5.14w+1.57 w=0\\3.57w =P\\w = \frac{P}{3.57} =0.2801P$

$l = \frac{P-2.57w}{2}\\l = 0.1401P$

Dimensions can be

$l=0.1401P\\w =0.2801P$

5 0

3 years ago

Factor the polynomial.

mrs_skeptik [129]

The coefficients here are 4, 32, and -24. The GCF of these values is 4.
As for the variables, the values are x^7, x^5, and x^4. The GCF is x^4
To factor this expression take out 4x^4 from each of the terms.
Coefficients should be divided by four and exponents should be subtracted by 4 since they have the same base (x).
The answer here is A.

5 0

3 years ago

How do you solve this

marusya05 [52]

Hope this helps! Good luck!

3 0

3 years ago

The house number if two adjacent homes are two consecutive even numbers. If their sum is 406, find the house numbers.

USPshnik [31]

Answer: follow my Instagram rihannah0423

Step-by-step explanation: PERIODT!!!!

5 0

3 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

7 0

3 years ago