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Dimas [21]
3 years ago
6

PLease help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
kirza4 [7]3 years ago
6 0

Answer:

2vt

Step-by-step explanation:

bro i can't answer now but i will answer it later i am randomly typing because to save it sorry but i will answer your question .

Mamont248 [21]3 years ago
4 0

Answer: 3t-12v+24

Step-by-step explanation:

1. Multiply 3 to everything inside the parentheses

3t +3(-4v)+3(8)

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It’s would be the second one aka it would be b
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A car driving at 10.0 m/s southward accelerates at 4.11 m/s2 southward for 7.25 s. what is the car's speed after this amount of
Anon25 [30]

The car's speed after the given amout of time is 19.8 m/s²

<h3>How would you define acceleration?</h3>

The rate at which an object's velocity with respect to time changes is referred to as acceleration in mechanics. They are vector quantities, accelerations. The direction of the net force acting on an object determines the direction of its acceleration. A point or object going straight ahead is accelerated when it accelerates or decelerates. Any alteration in an object's velocity could take the form of a change in direction of motion or a speed increase or reduction. The moon orbiting the earth, an apple falling to the ground, and a car stopped at a stop sign are a few instances of acceleration.

Average Acceleration =Change in Velocity/ Time Taken

So Change in Velocity is 10-x m/s

Time taken is 7.25 s

Given accelaration 4.11 m/s²

Putting the values we get,

4.11 = (10-x)/7.25

⇒29.8 = 10-x

⇒x=19.8 m/s²

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To learn more about acceleration, click here:

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4 0
2 years ago
HELP PLEASE!!!!! Solve each system.
Ede4ka [16]

15.\ \text{Elimination method}\\\underline{+\left\{\begin{array}{ccc}y=-3x+9\\y=3x+3\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2y=12\qquad\text{divide both sides by 2}\\.\qquad \boxed{y=6}\\\\\text{put the value of y to the second equation}\\\\6=3x+3\qquad\text{subtract 3 from both sides}\\3=3x\qquad\text{divide both sides by 3}\\1=x\to\boxed{x=1}\\\\Answer:\ x=1,\ y=6\to(1,\ 6)


16.\ \text{Substitution method}\\\left\{\begin{array}{ccc}y=11x+4\\y=7x-16\end{array}\right\Rightarrow11x+4=7x-16\\\\11x+4=7x-16\qquad\text{subtract 4 from both sides}\\11x=7x-20\qquad\text{subtract 7x from both sides}\\4x=-20\qquad\text{divide both sides by 4}\\\boxed{x=-5}\\\\\text{Put the value of x to the first equation}\\\\y=11(-5)+4\\y=-55+4\\\boxed{y=-51}\\\\Answer:\ x=-5,\ y=-51\to(-5,\ -51)

7 0
4 years ago
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