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PolarNik [594]
3 years ago
13

The distances (y), in miles, of two cars from their starting points at certain times (x), in hours, are shown by the equations b

elow:
Car A
y = 60x + 10

Car B
y = 40x + 70

After how many hours will the two cars be at the same distance from their starting point and what will that distance be?


a. 2 hours, 150 miles
b. 2 hours, 190 miles
c. 3 hours 150 miles
d. 3 hours, 190 miles
Mathematics
2 answers:
Zinaida [17]3 years ago
7 0

Answer:

The correct option is d.

Step-by-step explanation:

The distances (y), in miles, of two cars from their starting points at certain times (x), in hours, are shown by the equations

Car A:

y=60x+10             ... (1)

Car B:

y=40x+70            .... (2)

Equate equation (1) and (2), to find the hours after which the two cars be at the same distance from their starting point.

60x+10=40x+70

60x-40x=70-10

20x=60

x=3

The value of x is 3. It means after 3 hours two cars be at the same distance from their starting point.

Substitute x=3 in equation (1) to find the distance.

y=60(3)+10

y=180+10=190

The distance is 190 miles.

Therefore option d is correct.

Gnoma [55]3 years ago
3 0
We can simply set the two equations equal to each other:

60x+10=40x+70

We subtract 40x+10 from both sides to get:

20x=60

Divide by 20 to find that x=3.  Substitute this back into either equation to find the amount of miles - using the first, 60*3+10=190, so the answer is D.
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In a G.P the difference between the 1st and 5th term is 150, and the difference between the
liubo4ka [24]

Answer:

Either \displaystyle \frac{-1522}{\sqrt{41}} (approximately -238) or \displaystyle \frac{1522}{\sqrt{41}} (approximately 238.)

Step-by-step explanation:

Let a denote the first term of this geometric series, and let r denote the common ratio of this geometric series.

The first five terms of this series would be:

  • a,
  • a\cdot r,
  • a \cdot r^2,
  • a \cdot r^3,
  • a \cdot r^4.

First equation:

a\, r^4 - a = 150.

Second equation:

a\, r^3 - a\, r = 48.

Rewrite and simplify the first equation.

\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}.

Therefore, the first equation becomes:

a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150..

Similarly, rewrite and simplify the second equation:

\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}.

Therefore, the second equation becomes:

a\, r\, \left(r^2 - 1\right) = 48.

Take the quotient between these two equations:

\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}.

Simplify and solve for r:

\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}.

8\, r^2 - 25\, r + 8 = 0.

Either \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16} or \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}.

Assume that \displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}.

Similarly, assume that \displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}. Substitute back to either of the two original equations to show that \displaystyle a = \frac{497\, \sqrt{41}}{41} - 75.

Calculate the sum of the first five terms:

\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}.

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