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SCORPION-xisa [38]
1 year ago
13

When an object is dropped from a building (height 1450 feet )

Mathematics
1 answer:
Law Incorporation [45]1 year ago
7 0

Given

h=P(t)=-16t^2+1450

Evaluate P(5), as shown below

\begin{gathered} P(5)=-16*5^2+1450 \\ \Rightarrow P(5)=-400+1450=1050 \end{gathered}

Therefore, the answer is 1050ft.

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skad [1K]
<span>8d – 4d – 6d – 8 = 2d

Combine common terms:
-2d - 8 = 2d

Add 2d to both sides:
-8 = 4d

Swtich sides:
4d = -8

Divide both sides by 4:
d = -2

Answer: - 2
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3 0
3 years ago
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How many rows and colums in 16 divided by 112
earnstyle [38]
The answer to this question is:0.1428571429.
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4 0
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(2*)2-3×2*+2=0<br>4m-15°(×)m+75°​
Valentin [98]

Answer:

First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z

=

1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

7 0
2 years ago
The debate team has 15 members. Each member sold c coupon books as a fundraiser. Write an expression that shows the number of co
Svetllana [295]

15c

The team has fifteen members and each member sells an unknown amount of coupon books.

7 0
3 years ago
A population p of migrating butterflies changes over time it is represented by the equation p=100,000*4/5 where w is the number
Trava [24]

Given that,

A population p of migrating butterflies changes over time it is represented by the equation

p=100,000\times (\dfrac{4}{5})^w Where w is number of weeks.

To find,

The population after 2 weeks.

Solution,

We have,

p=100,000\times (\dfrac{4}{5})^w

Put w = 2 in the above equation.

p=100,000\times (\dfrac{4}{5})^2\\\\=100,000\times \dfrac{16}{25}\\\\=64000

So, the population after 2 weeks is 64000 .

7 0
3 years ago
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