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grigory [225]
3 years ago
13

Hey u all what up u all

Mathematics
2 answers:
Arlecino [84]3 years ago
8 0

Answer:

not too much, you?

Step-by-step explanation:

andrew11 [14]3 years ago
3 0

Well , good ..but I’m still waiting for someone to answer my question so..ye

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pishuonlain [190]

Answer: 2x + 5y = -5

Step-by-step explanation:

Two lines are said to be parallel if they have the same slope.

The equation of the line given :

2x + 5y = 10

To find the slope , we will write it in the form y = mx + c , where m is the slope and c is the y - intercept.

2x + 5y = 10

5y = -2x + 10

y = -2/5x + 10/5

y = -2/5x + 2

This means that the slope is -2/5 ,the line that is parallel to this line will also have a slope of -2/5.

using the formula:

y-y_{1} = m (x-x_{1} ) to find the equation of the line , we have

y - 1 = -2/5(x -{-5})

y - 1 = -2/5 ( x + 5 )

5y - 5 = -2 ( x + 5 )

5y - 5 = -2x - 10

5y + 2x = -10 + 5

therefore :

2x + 5y = -5 is the equation of the line that is parallel to 2x + 5y = 10

8 0
2 years ago
Read 2 more answers
You bought a video game for $35. You got it on sale
never [62]

Answer:

Sorry I feel bad I don't kt the answer sorry hope you find the answer tho

Step-by-step explanation:

sorry

7 0
3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
Help me plz it’s important
andre [41]
Is it Im1,im2, 1m3??
7 0
2 years ago
Can someone help me with this ​
jarptica [38.1K]
The answer is y=- 7/6x+8
Plz mark me Brainly
6 0
2 years ago
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