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Nuetrik [128]
3 years ago
14

What is the equation of a line passing through (-3, 7) and having a slope of -1/5 ? A. y = -5x + 22 B. y = 5x + 22 C. y=1/5x+32/

5 D. y=-1/5x+32/5 E. y=-1/5x+22
Mathematics
1 answer:
bogdanovich [222]3 years ago
5 0

Answer:

  D.  y = -1/5x +32/5

Step-by-step explanation:

The equation of the line passing through point (h, k) with slope m can be written ...

  y = m(x -h) +k

Filling in your given values, the equation is ...

  y = -1/5(x -(-3)) +7

  y = -1/5x -3/5 +7 . . . . eliminate parentheses

  y = -1/5x +32/5 . . . . . combine constants

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Find 1st, 2nd, 3rd, 4th and 10th nTh term. rule is 3n+4
Irina-Kira [14]

Answer:

When n is 1

3n+4

=3*1+4

=3+4

=7

When n is 2

3n+4

=3*2+4

=6+4

=10

When n is 3

3n+4

=3*3+4

=9+4

=13

When n is 4

3n +4

=3*4+4

=12+4

=16

When n is 10

3n+4

=3*10+4

=30+4

34

7 0
3 years ago
Jace walked a total distance of StartFraction 5 Over 6 EndFraction of a mile to school this week. If he walked the same distance
Nezavi [6.7K]
Its A 1/6

I looked it up!
8 0
2 years ago
Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p
GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2

you replace t1 from the first equation in the second equation:

(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}

Then, for t2 = 0 and t2=1/2 you obtain for t1:

t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths  intersect are:

r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)

B) You have the following two trajectories of two independent particles:

r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t

From the first equation you have:

t=1+2t\\\\t=-1

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0
3 years ago
Help please need this answer
fredd [130]

Answer:

(5,4) is the solution

Step-by-step explanation:

2x + y = 14 --> y = -2x + 14

Substitute y = -2x + 14 into 3x - 2y = 7


3x - 2( -2x + 14) = 7

3x + 4x - 28 = 7

7x = 35

x = 5

y = -2(5) + 14

y = -10 + 14

y = 4

Answer (5 , 4)

7 0
3 years ago
Read 2 more answers
If I worked all these hours, how much should I be paid. It's 13.69 per hour btw each. Schedule:
Dennis_Churaev [7]

Answer:

$246.42

Step-by-step explanation:

13.69(4+4+4+4) + 27.38

    13.69(16) + 27.38

     219.04 + 27.38

         246.42

You should make $246.42.

5 0
3 years ago
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