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morpeh [17]
3 years ago
13

Choose the equation that represents this situation. Use n to represent the

Mathematics
2 answers:
Flauer [41]3 years ago
5 0

Answer:

C

Step-by-step explanation:

I know is C let em have brainliest

Alenkasestr [34]3 years ago
4 0

Answer:

I THINK ITS C

Step-by-step explanation:

You might be interested in
If a third of the students entering high school drop out before graduating, but three-fifths eventually return and get their dip
GrogVix [38]

Answer:

\dfrac{13}{15}

Step-by-step explanation:

If a third drop out, Initial fraction of those who graduate = 1-\frac{1}{3} =\frac{2}{3}

Three-fifths of those who dropped out(one third) eventually return                         = \frac{3}{5} of \frac{1}{3} =\frac{3}{5} X \frac{1}{3}=\frac{1}{5}

The proportion of students entering high school who eventually graduate

= Those who graduated Initially+Those who returned

=\frac{2}{3}+\frac{1}{5}=\frac{10+3}{15}=\frac{13}{15}

7 0
3 years ago
Line segment AB has endpoints A(1, 4) and B(2, 8) . A dilation, centered at the origin, is applied to AB¯¯¯¯¯ . The image has en
kumpel [21]
To determine the scale factor of the dilation, we determine the distances between the endpoints of the two lines through the equation,
                            d = √(x₂ - x₁)² + (y₂ - y₁)²
 Substituting the known values.

Line Segment 1:     d = √(1 - 2)² + (4 - 8)²  = 4.123

Line Segment 2:     d = √(18 - 14)² + (12 - 1)² = 11.70

Dividing the answers will give us 0.352
7 0
3 years ago
Read 2 more answers
Two particles move in the xy-plane. At time t, the position of particle A is given by x(t)=5t−5 and y(t)=2t−k, and the position
Ipatiy [6.2K]

Answer:

Part A)  Not collide

Part B)  k = 4

Part C)  Particle B is moving fast.

Step-by-step explanation:

Two particles move in the xy-plane. At time, t

<u><em>Position of particle A:-</em></u>

x(t)=5t-5

y(t)=2t-k

<em><u>Position of particles B:-</u></em>

x(t)=4t

y(t)=t^2-2t+1

Part A)  For k = -6

Position particle A, (5t-5,2t+6)

Position of particle B, (4t,t^2-2t-1)

If both collides then x and y coordinate must be same

Therefore,

  • For x-coordinate:

5t - 5 = 4t    

       t = 5

  • For y-coordinate:

2t+6=t^2-2t-1

t^2-4t-7=0

t=-1.3,5.3

The value of t is not same. So, k = -6 A and B will not collide.

Part B) If both collides then x and y coordinate must be same

  • For x-coordinate:

5t - 5 = 4t    

       t = 5

  • For y-coordinate:

2t-k=t^2-2t-1

Put t = 5

10-k=25-10-1

k=4

Hence, if k = 4 then A and B collide.

Part C)

Speed of particle A, \dfrac{dA}{dt}

\dfrac{dA}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}

\dfrac{dA}{dt}=2\cdot \dfrac{1}{5}\approx 0.4

Speed of particle B, \dfrac{dB}{dt}

\dfrac{dB}{dt}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}

\dfrac{dB}{dt}=2t-2\cdot \dfrac{1}{4}

At t = 5

\dfrac{dB}{dt}=10-2\cdot \dfrac{1}{4}=2

Hence, Particle B moves faster than particle A

8 0
4 years ago
1.
Free_Kalibri [48]
1. 0.01333
2. 0.29069767
3. 0.45977011
7 0
3 years ago
Use Pascal’s Triangle to determine the fifth term of the expansion of (x − 5)6.
Alenkasestr [34]

Answer:

I need to see the choices for it

Step-by-step explanation:

6 0
3 years ago
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