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andrew-mc [135]

2 years ago

5

assuming a trucker changes his tires every for 40,000 miles and that he starts with a brand new set of tires how many sets of ti

res will be used in a year

1 answer:

sdas [7]2 years ago

8 0

80,000 tires will be used

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Please Help me with this question

nevsk [136]

Answer:

oe hello im from nepal follow me plz bro

3 0

3 years ago

Which numbers are the extremes of the proportion shown below?

wolverine [178]

Answer:

A 3 and 8

Step-by-step explanation:

if a/b=c/d, the extremes are "a" and "d". The means would then be "b" and "c"

8 0

2 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

2 years ago

Tell whether x and y show direct variation.Explain your reasoning.If so, find k #3

WINSTONCH [101]

Answer:

x and y do not show direct variation.

Step-by-step explanation:

The formula for direct variation is

y = kx Divide each side by y

k = y/x

k should have the same value for every point except, of course, (0,0)

For the first point,

k = 2/(-1) = -2

For the third point,

k = 2/1 = 2

The values of k are different, so x and y <em>do not</em> show direct variation.

7 0

3 years ago

Read 2 more answers

The population of a small town in central Florida has shown a linear decline in the years 2001-2012. In 2001 the population was

earnstyle [38]

Answer:

Step-by-step explanation:

First of all, to keep our numbers manageable, we are going to let year 2001 = 0 so year 2012 = 11. The coordinates that result from these years/population numbers are (0, 22200) and (11, 13950). Since this linear, we can plug those 2 coordinate pairs into the slope equation to find the rate at which the population is decreasing in people per year:

$m=\frac{13950-22200}{11-0}=\frac{-8250}{11}=-750\frac{people}{year}$ That means that the town is losing people at the rate of 750 per year. We can use that slop along with one of the coordinate points to write the linear equation representing this situation:

$y-22200=-750(x-0)$ and

y = -750x + 22200. That answers part A.

Now for B we need to find y when x = 13 (remember we let year 2001 = 0, so year 2014 = 13):

y = -750(13) + 22200 so

y = 12450 people in the year 2014

6 0

2 years ago