Which of the following products will result in an answer less than zero?

2 answers:

8 0

First, it is important to understand what happens to the sign of a product of two numbers depending on whether the two numbers are positive or negative. If you multiply two positive numbers, the product is positive; if you multiply a positive number and a negative number, the product is negative; if you multiply two negative numbers, the product is positive.

Since the first three choices would all result in a positive product, the correct choice is (-4)(7) because it would result in -28 which is below zero.

lidiya [134]3 years ago

7 0

Answer: (-4)(7)

Step-by-step explanation:

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Marysya12 [62]

The correct answer would be 48 min. (Choice B)

In an inverse relationship, we multiply the 2 variables together to get a constant.

Our equation would be:

GT = C If we plug in the numbers we have:

8(12) = 96

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Please help me with the below question.

o-na [289]

Given that

$\vec r(t) = \left\langle 4t e^{-t}, 8 \arctan(t), 8 e^t \right\rangle$

we first differentiate with respect to t to get the tangent vector, $\vec T(t)$ :

$\vec T(t) = \dfrac{d\vec r}{dt} = \left\langle (4-4t) e^{-t}, \dfrac8{1+t^2}, 8e^t \right\rangle$

At t = 0, the tangent vector is

$\vec T(0) = \left\langle 4, 8, 8 \right\rangle$

To get the unit tangent vector, multiply this by 1/(norm of tangent vector) :

$\|\vec T(0)\| = \sqrt{4^2 + 8^2 + 8^2} = \sqrt{144} = 12$

Then the unit tangent vector is

$\dfrac{\vec T(0)}{\|\vec T(0)\|} = \left\langle \dfrac4{12}, \dfrac8{12}, \dfrac8{12}\right\rangle = \boxed{\left\langle\dfrac13, \dfrac23, \dfrac23\right\rangle}$

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An object is dropped from a small plane. As the object falls, its distance, d, above the ground after t seconds, is given by the

DedPeter [7]

The inequality can be used to find the interval of time taken by the object to reach the height greater than 300 feet above the ground is:

$d = -16t^2 + 1000$

Solution:

The object falls, its distance, d, above the ground after t seconds, is given by the formula:

$d = -16t^2 + 1000$

To find the time interval in which the object is at a height greater than 300 ft

Frame a inequality,

$-16t^2 + 1000 > 300$

Solve the inequality

Subtract 1000 from both sides

$-16t^2 + 1000 - 1000 > 300 - 1000\\\\-16t^2 > -700$

$16t^2 < 700\\\\Divide\ both\ sides\ by\ 16\\\\t^2 < \frac{700}{16}\\\\Take\ square\ root\ on\ both\ sides\\\\t < \sqrt{\frac{700}{16}}\\\\t < \pm 6.61$