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crimeas [40]
3 years ago
14

Two cars travel the same distance. The first car travels 40 mph and reaches its destination in t hours.

Mathematics
1 answer:
Vesna [10]3 years ago
3 0

Answer:

<em>The first car reaches its destination at 11 hours</em>

Step-by-step explanation:

<u>Distance Traveled </u>

The distance X traveled by and object moving at constant speed is given by:

X=v.t

Where v is the speed and t is the time.

We are given some input data. Two cars travel the same distance, let's call it X. The first car travels at 40 mph and reaches its destination in t hours.

We also know the second car travels at 55 mph and reaches the destination 3 hours earlier. The time used by the second car is t-3 hours

The equation for the distance of the first car is

X=40t

The equation for the distance of the second car is

X=55(t-3)

Both distances are the same, thus

40.t=55(t-3)

Operating:

40t=55t-165

Joining like terms:

15t=165

Solving:

t=165/15

t=11 hours

The first car reaches its destination at 11 hours.

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Five times the sum of a number and 27 is greater than or equal to six times the sum of that
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The complete version of question:

<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.</em>

Answer:

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

Step-by-step explanation:

As the description of the statement is:

'<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.</em>

<em />

As

  • <em>Five times the sum of a number and 27  </em>is written as: 5(x + 27)
  • <em>greater than or equal </em>is written as:  \geq
  • <em>six times the sum of that number and 26'  </em>is written as: 6(x + 26)

so lets combine the whole statement:

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

solving

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

<em />5x+135\ge \:6x+156<em />

<em />5x+135-135\ge \:6x+156-135<em />

<em />5x\ge \:6x+21<em />

<em />5x-6x\ge \:6x+21-6x<em />

<em />-x\ge \:21<em />

<em />\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}<em />

<em />\left(-x\right)\left(-1\right)\le \:21\left(-1\right)<em />

<em />x\le \:-21<em />

Therefore,

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

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The answer is c you must look at key words i learned my key words

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