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3 years ago

What are the answers to the "Writing linear equations puzzle"? Plz help!

Mathematics

1 answer:

saveliy_v [14]3 years ago

6 0

Step-by-step explanation:

Can you put a picture or give more information?

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Simplify the Expression<br> 2(x +14) - (2x - 5)

elena55 [62]

your answer would be 33.

step 1

2(+14)−1(2−5)

2+28−1(2−5)

step 2 distribute

2+28−1(2−5)

2+28−2+5

step 3 then add the numbers

2+28−2+5

2+33−2

combine

2+33−2

ANSWER IS

7 0

3 years ago

A spaceship is traveling at a steady rate. The table shows the distance the spaceship travels in a given period of time. Find th

nika2105 [10]

The distance of the spaceship in discuss as in the task content given can be evaluated as; 800miles.

<h3>What is the distance the spaceship travels in 4 minutes?</h3>

The distance travelled by the spaceship in discuss can be evaluated by means of the slope of the linear relationship as follows;

Hence it follows from ratios that by observation, the linear relationship has a slope of 200mi/min.

Consequently, we can evaluate the distance travelled after 4 minutes as;

Distance = 200 × 4 = 800mi.

Ultimately, the distance travelled per minute by the spaceship is; 800mi.

Remarks:

600 miles

520 miles

800 miles

1,080 miles

Read more on ratios;

brainly.com/question/13513438

#SPJ1

4 0

1 year ago

Find the equation of a straight line R, which passes through (50, 2) and is parallel to the line y + 10 = 2

Snowcat [4.5K]

Answer:

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and so sorry I don't no answer

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6 0

3 years ago

What is the solution of the system of equations? y = –4x + 10 y = –2x – 6

Elan Coil [88]

D. (8, -22) is the correct answer

4 0

3 years ago

1. The position of a particle moving along a coordinate axis is given by: s(t) = t^2 - 5t + 1. a) Find the speed of the particle

zimovet [89]

Answer: $\left | 2t-5\right |,\ 2,\ 2t-5$

Step-by-step explanation:

Given

Position of the particle moving along the coordinate axis is given by

$s(t)=t^2-5t+1$

Speed of the particle is given by

$\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=\left | 2t-5\right |$

Acceleration of the particle is

$\Rightarrow a=\dfrac{dv}{dt}\\\\\Rightarrow a=2$

velocity can be negative, but speed cannot

$\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=2t-5$

3 0

3 years ago