Statement: If p, then q.

Find the slope Options B.-1 C.1/4 D.4

Ber [7]

<h3>Answer: D) 4</h3>

Explanation:

Start at the point (-1,-4). In order to move to (0,0), we must do the following two steps:

Move 4 units up
Move 1 unit to the right

This indicates slope = rise/run = 4/1 = 4.

You could use the slope formula to get this same result.

5 0

3 years ago

The submarine is currently located at 10,000 feet below sea level but then rises 5,200. Where is the

Ilia_Sergeevich [38]

Answer:

A. 4,800 feet below sea level

Step-by-step explanation:

10,000 feet below sea level = -10,000 feet

"rises 5,200 feet" = +5,200 feet

=-10,000 + 5,200 = -4,800 feet

=4,800 feet below sea level.

8 0

3 years ago

A canister is dropped from a helicopter 500m above the ground. Its parachute does not open, but the canister has been designed t

solmaris [256]

Answer:

The impact speed 98.995 m/s is less than 100 m/s and the canister will not burst.

Step-by-step explanation:

A function F is called an antiderivative of f on an interval I if $F'(x) = f(x)$ for all x in I.

Recall that if the object has position function $s=f(t)$ , then the velocity function is $v(t)=s'(t)$ . This means that the position function is an antiderivative of the velocity function. Likewise, the acceleration function is $a(t)=v'(t)$ , so the velocity function is an antiderivative of the acceleration.

An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by $g$ . For motion close to the ground we may assume that $g$ is constant, its value being about $9.8 \:{\frac{m}{s^2}}$ .

We know that the acceleration due to gravity is given by

$a(t)=-9.8$

and the antiderivative is velocity

$v(t)=\int a(t)\,dt\\v(t)=\int -9.8\,dt\\v(t)=-9.8t +C$

We know that the canister was dropped, so the initial velocity at t = 0 is zero, this fact let us know the value of C.

$v(0)=9.8(0)+C\\C=0$

The antiderivative of velocity is the position

$s(t)=\int v(t) \, dt\\s(t)=\int -9.8t \, dt\\s(t)=-4.9t^2+C$

To find the value of the constant C, we know that the height was 500 m at t = 0, this means $s(0)=500$

$500=-4.9(0)^2+C\\C=500$

$s(t)=-4.9t^2+500$

Using the fact that at the time of impact the height s(t) is zero we can compute the total time of the fall:

$s(t)=-4.9t^2+500=0\\\\-4.9t^2=-500\\\\t^2=\frac{5000}{49}\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\t=\sqrt{\frac{5000}{49}},\:t=-\sqrt{\frac{5000}{49}}$