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Scorpion4ik [409]
3 years ago
13

Write an equation for a line that passes through the Irgun and perpendicular to y=5x-2

Mathematics
1 answer:
umka21 [38]3 years ago
7 0

Answer:

Well we have to find eq of line passing through origin and perpendicular to y=5x-2

First you need slope

As slope of y=5x-2 line is 5

So, slope of perpendicular line is -1/5

Also it passes through origin

as we know eq of line is y-y1=m(x-x1)

substitute X1,y1 = (0,0)

it becomes y =mx

so, y = -x/5

5y +x = 0 is required equation

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A bag contains 8 purple marbles, 7 yellow marbles, 8 red marbles, and 3 blue marbles. What is the probability of drawing a purpl
statuscvo [17]

Answer:

4:26

Step-by-step explanation:

so count all of the marbles but this time you need to take away  a purple marble replace it with a blue so there would be 4 blues in the bag.

5 0
3 years ago
Identify the solutions to the quadratic equation.<br> Check all that apply.<br> x2 + 4x - 21 = 0
Vika [28.1K]

Answer:

C and D

Step-by-step explanation:

(X-3)(X+7)=0

X-3=0   X+7=0

X=3  X=-7

8 0
3 years ago
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What’s the definition of percent equation
geniusboy [140]
I think the asnwet is 2 also
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
Simplify the following expression. 2^2 • 2^3
sesenic [268]

Answer:

4 × 6

Step-by-step explanation:

2^2= 4 or 2 × 2= 4

2^3= 6 or 2 × 2 × 2= 6

So, 4 × 6

8 0
3 years ago
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