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3 years ago

Can someone help me with this please I need help

Mathematics

2 answers:

LUCKY_DIMON [66]3 years ago

8 0

Answer:

This is linear.

Step-by-step explanation:

Its because it has a constant rate, no sudden changes.

likoan [24]3 years ago

7 0

Answer: Linear

Step-by-step explanation:

This is linear because when graphed it is straight.

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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

Position of particle A:-

$x(t)=5t-5$

$y(t)=2t-k$

Position of particles B:-

$x(t)=4t$

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

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

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3 years ago

Which object has a mass of approximately 450 grams?

Lostsunrise [7]

Football is your answer

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3 years ago

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In a group of eighteen people, each person shakes hands once with each other person in the group. How many handshakes will occur

miv72 [106K]

The number of handshakes that will occur in a group of eighteen people if each person shakes hands once with each other person in the group is 153 handshakes

In order to determine the number of handshakes that will occur among 18 people, that is, the number of ways we can choose 2 persons from 18 people.

∴ The number of handshakes = $^{18}C_{2}$