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Lena [83]
4 years ago
15

A particle moves on the x-axis with velocity given by v(t) = 3t4 − 11t2 + 9t − 2 for −3 ≤ t ≤ 3 . How many times does the partic

le change direction as t increases from to −3 to 3?
Mathematics
1 answer:
Natali [406]4 years ago
5 0

Answer:

The particle changes its direction 2 times within the time -3<t<3

Step-by-step explanation:

The particle is moving only in a single dimension (x-axis), and whenever the particle changes its direction it also means that it's velocity while changing the direction will be zero.

Hence,

v(t) = 0

but since we're not concerned with the actual values of t when v(t)=0, we'll only consider how many times does the particle changes its direction.

for that we'll simply plot the curve using half-steps from -3 to 3.

t, v(t)

-3, 115

-2.5, 23.9375

-2, -16

-1.5, -25.0625

-1, -19

-0.5, -9.0625

0, -2

0.5, -0.0625

1, -1

1.5, 1.9375

2, 20

2.5, 68.9375

3, 169

What we need to check is at what points does the sign of v(t) values change (because only between those points will v(t) cross the x-axis, hence it's value would've crossed 0)

so there are two points!

between the intervals t = [-2.5,2] and [1,1.5]

so there are two points where the particle changes its directions and those points lie somewhere between these two aforementioned intervals.

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Step-by-step explanation:

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Help me out please. For 20 points.
Nata [24]

Answer:

7 1/3

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
6x+2(x+4)&lt;2x+20 solve for inequality
8_murik_8 [283]

The problem to solve is:

6x+2(x+4)‹2x+20

First, let's work on the left hand side of your inequality, the 6x+2(x+4)

This means, for instance, to see if it can be simplified at all.

Multiply x and 6

Multiply x and 1

The x just gets copied along.

The answer is x

x

6*x evaluates to 6x

x+4 evaluates to x+4

Multiply 2 by x+4

we multiply 2 by each term in x+4 term by term.

This is the distributive property of multiplication.

Multiply 2 and x

Multiply 1 and x

The x just gets copied along.

x

2 × x = 2x

Multiply 2 and 4

1

2 × 4 = 8

2*(x+4) evaluates to 2x+8

6x + 2x = 8x

The answer is 8x+8

6*x+2*(x+4) evaluates to 8x+8

So, all-in-all, the left hand side of your inequality can be written as: 8x+8

Now, let's work on the right hand side of your inequality, the 2x+20

Multiply x and 2

Multiply x and 1

The x just gets copied along.

The answer is x

x

2*x evaluates to 2x

2*x+20 evaluates to 2x+20

The right hand side of your inequality can be written as: 2x+20

So with these (any) simplifications, the inequality we'll set out to solve is:

8x+8 ‹ 2x+20

Move the 8 to the right hand side by subtracting 8 from both sides, like this:

From the left hand side:

8 - 8 = 0

The answer is 8x

From the right hand side:

20 - 8 = 12

The answer is 12+2x

Now, the inequality reads:

8x ‹ 12+2x

Move the 2x to the left hand side by subtracting 2x from both sides, like this:

From the left hand side:

8x - 2x = 6x

The answer is 6x

From the right hand side:

2x - 2x = 0

The answer is 12

Now, the inequality reads:

6x ‹ 12

To isolate the x, we have to divide both sides of the inequality by the other "stuff" (variables or coefficients)

around the x on the left side of the inequality.

The last step is to divide both sides of the inequality by 6 like this:

To divide x by 1

The x just gets copied along in the numerator.

The answer is x

6x ÷ 6 = x

12 ÷ 6 = 2

The solution to your inequality is:

x ‹ 2

So, your solution is:

x must be less than 2The problem to solve is:

6x+2(x+4)‹2x+20

First, let's work on the left hand side of your inequality, the 6x+2(x+4)

This means, for instance, to see if it can be simplified at all.

Multiply x and 6

Multiply x and 1

The x just gets copied along.

The answer is x

x

6*x evaluates to 6x

x+4 evaluates to x+4

Multiply 2 by x+4

we multiply 2 by each term in x+4 term by term.

This is the distributive property of multiplication.

Multiply 2 and x

Multiply 1 and x

The x just gets copied along.

x

2 × x = 2x

Multiply 2 and 4

1

2 × 4 = 8

2*(x+4) evaluates to 2x+8

6x + 2x = 8x

The answer is 8x+8

6*x+2*(x+4) evaluates to 8x+8

So, all-in-all, the left hand side of your inequality can be written as: 8x+8

Now, let's work on the right hand side of your inequality, the 2x+20

Multiply x and 2

Multiply x and 1

The x just gets copied along.

The answer is x

x

2*x evaluates to 2x

2*x+20 evaluates to 2x+20

The right hand side of your inequality can be written as: 2x+20

So with these (any) simplifications, the inequality we'll set out to solve is:

8x+8 ‹ 2x+20

Move the 8 to the right hand side by subtracting 8 from both sides, like this:

From the left hand side:

8 - 8 = 0

The answer is 8x

From the right hand side:

20 - 8 = 12

The answer is 12+2x

Now, the inequality reads:

8x ‹ 12+2x

Move the 2x to the left hand side by subtracting 2x from both sides, like this:

From the left hand side:

8x - 2x = 6x

The answer is 6x

From the right hand side:

2x - 2x = 0

The answer is 12

Now, the inequality reads:

6x ‹ 12

To isolate the x, we have to divide both sides of the inequality by the other "stuff" (variables or coefficients)

around the x on the left side of the inequality.

The last step is to divide both sides of the inequality by 6 like this:

To divide x by 1

The x just gets copied along in the numerator.

The answer is x

6x ÷ 6 = x

12 ÷ 6 = 2

The solution to your inequality is:

x ‹ 2

So, your solution is:

x must be less than 2

7 0
3 years ago
I NEED HELP WILL GIVE BRAINLIEST!! PLEASE HELP ME ASAP!!!!
GenaCL600 [577]

Answer:

A. The length of the second leg is 8.5 inches

B. The length of the three-dimensional diagonal is 9.9 inches

Step-by-step explanation:

Let us revise the relation between the hypotenuse and the two legs of a right triangle

(hypotenuse)² = (vertical leg)² + (horizontal leg)²

∵ The length of the rectangular box = 8 inches

∵ The width of the rectangular box = 3 inches

∵ The height of the rectangular box = 5 inches

∵ Length and width are perpendicular to each other

∴ The Δ whose legs are 3 and 8 is a right triangle

In the right Δ whose legs are 3 and 8

∵ (hypotenuse)² = (3)² + (8)²

∴ (hypotenuse)² = 9 + 64

∴ (hypotenuse)² = 73

- Take √  for both sides

∴ hypotenuse = \sqrt{73} = 8.544003745

- Round it to the nearest tenth of one inch

∴ hypotenuse = 8.5 inches

A.

The 3-dimensional diagonal is the hypotenuse of a right triangle whose legs are the vertical edge and the hypotenuse of the right triangle whose legs are 3 and 8

∵ The hypotenuse of the right triangle whose legs are 3 and

   8 is 8.5 inches

∴ The length of the second leg is 8.5 inches

B.

In the right triangle whose hypotenuse is the 3-dimensional diagonal and legs are the vertical edge , the hypotenuse of the right triangle whose legs are 3 and 8

∵ (3-dimensional diagonal)² = (5)² + (73)²

∴ (3-dimensional diagonal)² = 25 + 73

∴ (3-dimensional diagonal)² = 98

- Take √ for both sides

∴ 3-dimensional diagonal = \sqrt{98} = 9.899494937

- Round it to the nearest tenth of an inch

∴ 3-dimensional diagonal = 9.9 inches

∴ The length of the three-dimensional diagonal is 9.9 inches

<em>V.I.N: you can find the length of the  three-dimensional diagonal by using this rule → </em>d=\sqrt{l^{2}+w^{2}+h^{2}}<em> </em>

3 0
4 years ago
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