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2 years ago

Evaluate -1 - [-z] where z = -2

Mathematics

2 answers:

slega [8]2 years ago

5 0

$\huge\textbf{Hey there!}$

$\huge\textsf{-1 - [-z]}\\\\\huge\textsf{= -1 - [- -2]}\\\\\huge\textsf{= -1 - - 2}\\\\\huge\textsf{= -1 + (-2)}\\\\\huge\textsf{= -1 - 2}\\\\\huge\textsf{= -3}\\\\\\\huge\textbf{Therefore, your answer is: \boxed{\mathsf{-3}}}\huge\checkmark$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

mixer [17]2 years ago

3 0

Answer:

-3

Step-by-step explanation:

Evaluate - 1 - ( - z ) where z = -2

you can find this out because when there are two minuses by each other it equals a positive but then z = -2 so that adds another minus so it narrows it down to -1+-2 which equals 3

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Step-by-step explanation:

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3 years ago

A sample of 400 grams of radioactive substance decays according to the function A(t) = 400e^-.046t, where t is the time in years

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Step-by-step explanation:

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4 years ago

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Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

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.

6 0

3 years ago