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3 years ago

1. Select all the equations that describe each situation and then find the solution,

Mathematics

1 answer:

Studentka2010 [4]3 years ago

7 0

Answer:iii.

Step-by-step explanation: you would take 14 and subtract 3 so it would look like this *14-3* and that is how you get your answer

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A car traveling east at 40.0 m/s passes a trooper hiding at the roadside. The driver uniformly reduces his speed to 25.0 m/s in

Anni [7]

Step-by-step explanation:

It is given that,

Initial speed of the car, u = 40 m/s

Final speed of the car, v = 25 m/s

Time taken, t = 3.5 s

(a) We need to find the magnitude and direction of the car’s acceleration as it slows down. It can be calculated using formula as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{25-40}{3.5}$

$a=-4.28\ m/s^2$

The acceleration is in the opposite direction of motion i.e. west.

(b) Let s is the distance the car travel in the 3.5-s time period. It can be calculated using the third equation of motion as

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{25^2-40^2}{2\times (-4.28)}$

s = 113.9 meters

Hence, this is the required solution.

6 0

3 years ago

What is -15x=0 solution

cricket20 [7]

Here is the method for solving that equation:

Step #1: Write the equation:

-15x = 0

Step #2: Divide each side of the equation by -15 :

<u>x = 0</u>

3 0

3 years ago

Find the value of each variable:

Leona [35]

Answer:

5√3 = b

5 = d

10√3 = a

15 = c

Step-by-step explanation:

Sin60° = opp/hyp

sin60° = b/10

10sin60° = b

5√3 = b

Cos60° = adj/hyp

Cos60° = d/10

10Cos60° = d

5 = d

Sin30° = opp/hyp

Sin30° = 5√3/a

a = 5√3/Sin30°

a = 10√3

Tan30° = opp/adj

Tan30° = 5√3/c

c = 5√3/tan30°

c = 15

3 0

3 years ago

During a cold spell in January 1989, Homer, Alaska, recorded

Oksanka [162]

I think the answer is -48°F

6 0

3 years ago

Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10

natka813 [3]

ANSWER

$( \frac{2}{3} , \frac{1}{5} )$

EXPLANATION

The first equation is

$\frac{1}{2} x - \frac{3}{4} y = \frac{11}{60} ...(1)$

The second equation is

$\frac{2}{5} x + \frac{1}{6} y = \frac{3}{10} ...(2)$

We want to eliminate y, so we multiply the first equation by

$\frac{4}{5}$

$\frac{4}{5} \times \frac{1}{2} x - \frac{4}{5} \times \frac{3}{4} y = \frac{11}{60} \times \frac{4}{5}$

$\frac{2}{5} x - \frac{3}{5} y = \frac{11}{75} ...(3)$

We now subtract equation (3) from (2)

$(\frac{2}{3} x - \frac{2}{3} x )+ ( \frac{1}{6} y - - \frac{3}{5}y ) =( \frac{3}{10} - \frac{11}{75} )$

$\frac{1}{6} y + \frac{3}{5}y =\frac{3}{10} - \frac{11}{75}$

$\frac{23}{30}y = \frac{23}{150}$

Multiply both sides by

$\frac{30}{23}$

$\implies \: \frac{30}{23} \times \frac{23}{30}y= \frac{23}{150} \times \frac{30}{23}$

$\implies \: y = \frac{1}{5}$

Substitute into the first equation to solve for x .

$\frac{1}{2} x - \frac{3}{4} \times \frac{1}{5} = \frac{11}{60}$

Multiply to obtain

$\frac{1}{2} x - \frac{3}{20} = \frac{11}{60}$

$\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}$

$\frac{1}{2} x = \frac{1}{3}$

Multiply both sides by 2.

$2 \times \frac{1}{2} x =2 \times \frac{1}{3}$

$x = \frac{2}{3}$

The solution is

$( \frac{2}{3} , \frac{1}{5} )$

5 0

3 years ago

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