Thanks for everyone who's helping :) please help for a grade and due today

Find the equation of the line that passes through (-1,-3) and is perpendicular to

miv72 [106K]

Answer:

y = -2x - 5

Step-by-step explanation:

2y = x - 3

y = 1/2x - 3/2

gradient = 1/2

perpendicular gradient = negative reciprocal = -2

y = -2x + c

(-3) = -2(-1) + c

-3 = 2 + c

-5 = c

3 0

2 years ago

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Please Help!

Flura [38]

Let x be the length of the train.

On the basis of the observer;
Speed of the train = x/6

On the basis of the bridge;
Total distance covered by any point of the train= 350+x
Speed = (350+x)/20

Equating the two expressions of speed;
x/6 = (350+x)/20
20(x) = 6(350+x)
20x = 2100+6x
(20-6)x = 2100
14x = 2100
x= 2100/14 = 150 m
Speed = x/6 = (350+x)/20 = 150/6 = 500/20 = 25 m/s.

Therefore,
Length of train = 150 m
Speed of train = 25 m/s

8 0

3 years ago

What is the nth term rule of the quadratic sequence below?

Vladimir [108]

Answer:

3n² + 5n - 2

Step-by-step explanation:

Given sequence:

6, 20, 40, 66, 98, 136, ...

Calculate the first differences between the terms:

$6 \underset{+14}{\longrightarrow} 20 \underset{+20}{\longrightarrow} 40 \underset{+26}{\longrightarrow} 66 \underset{+32}{\longrightarrow} 98 \underset{+38}{\longrightarrow} 136$

As the first differences are not the same, calculate the second differences:

$14 \underset{+6}{\longrightarrow} 20 \underset{+6}{\longrightarrow} 26 \underset{+6}{\longrightarrow} 32 \underset{+6}{\longrightarrow} 38$

As the second differences are the same, the sequence is quadratic and will contain an n² term.

The coefficient of the n² term is half of the second difference.

Therefore, the n² term is: 3n²

Compare 3n² with the given sequence:

$\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 & 3 & 12 & 27 & 48 \\\cline{1-5} \sf operation & +3&+8 & +13 & +18 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}$

The second operations are different, therefore calculate the differences between the second operations:

$3 \underset{+5}{\longrightarrow} 8 \underset{+5}{\longrightarrow} 13\underset{+5}{\longrightarrow} 18$

As the differences are the same, we need to add 5n as the second operation:

$\begin{array}{|c|c|c|c|c|}\cline{1-5} n & 1 & 2 & 3 & 4\\\cline{1-5} 3n^2 +5n & 8&22 & 42 & 68\\\cline{1-5}\sf operation & -2 &-2 &-2 & -2 \\\cline{1-5} \sf sequence & 6 & 20 & 40 & 66\\\cline{1-5}\end{array}$

Finally, we can clearly see that the operation to get from 3n² + 5n to the given sequence is to subtract 2.

Therefore, the nth term of the quadratic sequence is:

3n² + 5n - 2

6 0

1 year ago

Find the equation of the plane through the point p=(5,5,4)p=(5,5,4) and parallel to the plane 5x+2y−z=−6.

Anna35 [415]

The unit normal for the given plane is <5,2,-1>.
The equation of the plane parallel to the given plane passing through (5,5,4) is therefore
5(x-5)+2(y-5)-1(z-4)=0
simplify =>
5x+2y-z=25+10-4=31

Answer: the plane through (5,5,4) parallel to 5x+2y-z=-6 is 5x+2y-z=31

7 0

3 years ago

What shape is created by a cross-section of a triangular prism cut perpendicular to the bases?

Butoxors [25]

Answer:

Its a triangle .And the question says triangular

7 0

3 years ago