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3 years ago

Let f(x) = 13 - |2x+ 1|. Find all x for which f(x) less than or equal to 14

Mathematics

1 answer:

Art [367]3 years ago

8 0

Answer:

$x \leqslant -1\: or \: x \geqslant - 1$

Step-by-step explanation:

The given function is

$f(x) = 13 - |2x + 1|$

We want to find all values of x for which:

f(x) is greater than or equal to 14.

This implies;

$13 - |2x + 1| \geqslant 14$

Subtract 13 from both sides;

$- |2x + 1| \geqslant 14 - 13$

$|2x + 1| \geqslant - 1$

By definition of the absolute value function,

$- (2x + 1) \geqslant - 1 \: or \: (2x + 1) \geqslant - 1$

Divide through the first inequality and and reverse the inequality sign:

$2x + 1 \leqslant -1 \: or \: 2x + 1 \geqslant - 1$

$2x \leqslant -1 - 1 \: or \: 2x \geqslant - 1 - 1$

$2x \leqslant -2 \: or \: 2x \geqslant - 2$

$x \leqslant -1\: or \: x \geqslant - 1$

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A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra

Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

$z=1$

$\frac{dx}{dt}=430$

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y=2$

According to Pythagoras

$hypotenuse^2=base^2+perpendicular^2$

$y^2=x^2+1^2$

$y^2=x^2+1$ .... (1)

Put z=1 and y=2, to find the value of x.

$2^2=x^2+1^2$

$4=x^2+1$

$4-1=x^2$

$3=x^2$

Taking square root both sides.

$\sqrt{3}=x$

Differentiate equation (1) with respect to t.

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

Divide both sides by 2.

$y\frac{dy}{dt}=x\frac{dx}{dt}$

Put $x=\sqrt{3}$ , y=2, $\frac{dx}{dt}=430$ in the above equation.

$2\frac{dy}{dt}=\sqrt{3}(430)$

Divide both sides by 2.

$\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}$

$\frac{dy}{dt}=372.390923627$

$\frac{dy}{dt}\approx 372$

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

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Answer:

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Step-by-step explanation:

$\sf 3^2\times \:3^4\times \:3^6$

<u>Apply exponent rule:</u>

$\hookrightarrow \boxed{\sf a^b\times \:a^c=a^{b+c}}$

$\sf 3^2\times \:3^4\times \:3^6$

$\sf 3^{2+4+6}$

Add the numbers:

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$\sf \:3^{12}$

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