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2 years ago

Find the slope of the line

Mathematics

2 answers:

PolarNik [594]2 years ago

8 0

Answer:

Step-by-step explanation:

ask a teacher so people might lie abt the answer

Mama L [17]2 years ago

3 0

Answer:

- 5/3

Step-by-step explanation:

This slope would be a positive slope since we have the dot on the negative side but it looks like it's in the ones but it's negative 5/3 in my opinion good luck

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A scientist listed the volumes, in milliliters, of some liquids used in an experiment.

Aloiza [94]

Answer:

B) 1.078 < 1.23 (T)

Step-by-step explanation:

A) 1.23 < 1.203 (F)

B) 1.078 < 1.23 (T)

C) 1.17 > 1.203 (F)

D) 1.078 > 1.17 (F)

4 0

3 years ago

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Jamie is trying to work out the surface area of the cuboid below. Spot any mistakes and write the correct answer.

umka21 [38]

Step-by-step explanation:

The dimensions of the cuboid are 5 cm, 7 cm and 9 cm.

The formula for the surface area of the cuboid is as follows :

$A=2(lb+bh+hl)$

We have,

b = 9 cm, l = 7 cm and h = 5 cm

So,

$A=2(9\times 7+9\times 5+7\times 5)\\\\A=2\times 143\\\\A=286\ cm^2$

Jamie forgets to multiply 143 by 2 as the formula for surface area contains 2 as well.

7 0

3 years ago

Please help mehhh Thank you! :)

vivado [14]

Parentheses first -4+-8= -12(1)= -12
= -12

4 0

3 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$