Step Statement Reason

According to a state law for vehicles traveling on state roads, the maximum total weight of a vehicle and its contents depends o

Alex787 [66]

Inequalities are used to show the relationship between unequal expressions.

The inequalities that represent the maximum weights are:

2 axles - $x \le 40000$ .
3 axles - $x \le 60000$ .
4 axles - $x \le 80000$ .

Let the maximum weight be represented with x

In inequality, maximum means less than or equal to i.e. $\le$

(a) 2 axles

The maximum weight, here is 40000.

So, the inequality is:

$x \le 40000$

(b) 3 axles

The maximum weight, here is 60000.

So, the inequality is:

$x \le 60000$

(c) 4 axles

The maximum weight, here is 80000.

So, the inequality is:

$x \le 80000$

See attachment for the graphs of each inequality

Read more about inequalities at:

brainly.com/question/11612965

6 0

3 years ago

What's the lcm for 18, 27, 12? I got 648. Am not sure if it is correct. Can you please explain if it is correct or not?

erik [133]

Answer:

Step-by-step explanation:

Put these numbers into their prime factors.

18: 2 * 3 * 3

27: 3 * 3 * 3

12: 2 * 2 * 3

The LCM must have

two 2s

Three 3s

That's it

2*2 * 3 * 3 * 3

LCM = 108

648 might be a multiple of the three, but it's not the smallest one.

108 = 27 * 4

108 = 18 * 6

108 = 12 * 9

6 0

3 years ago

What is the solution for the following equation? x^2-6x+9=11

tankabanditka [31]

Make one side equal zero
minus 11 to both sides
x^2-6x-2=0
another quadratic equation

if you hahve
ax^2+bx+c=0
x= $\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$

1x^2-6x-2=0
a=1
b=-6
c=-2

x= $\frac{-(-6)+/- \sqrt{(-6)^{2}-4(1)(-2)} }{2(1)}$
x= $\frac{6+/- \sqrt{36+8} }{2}$
x= $\frac{6+/- \sqrt{44} }{2}$
x= $\frac{6+/- 2\sqrt{11} }{2}$
x= $3+/- \sqrt{11}$
x= $3+ \sqrt{11}$ or $3- \sqrt{11}$

aprox
x=6.31662 or -0.316625

4 0

3 years ago

A homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his pro

Anestetic [448]

Answer:

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Step-by-step explanation:

Given that, a homeowner plans to enclose a 200 square foot rectangle playground.

Let the width of the playground be y and the length of the playground be x which is the side along the boundary.

The perimeter of the playground is = 2(length +width)

=2(x+y) foot

The material costs $1 per foot.

Therefore total cost to give boundary of the play ground

=$[ 2(x+y)×1]

=$[2(x+y)]

But the neighbor will play one third of the side x foot.

So the neighbor will play $=\$(\frac13x)$

Now homeowner's total cost for the material is

$=\$[ 2(x+y)-\frac13x]$

$=\$[2x+2y-\frac13x]$

$=\$[2y+x+x-\frac13x]$

$=\$[2y+x+\frac{3x-x}{3}]$

$=\$[2y+x+\frac{2}{3}x]$

$=\$[2y+\frac53x]$

$\therefore C(x)=2y+\frac53x$

where C(x) is total cost of material in $.

Given that the area of the playground is 200.

We know that,

The area of a rectangle is =length×width

=xy square foot

∴xy=200

$\Rightarrow y=\frac{200}{x}$

Putting the value of y in C(x)

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

The domain of C is $(0,\infty )$ .

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

Differentiating with respect to x

$C'(x)= - \frac{400}{x^2}+\frac53$

Again differentiating with respect to x

$C''(x) = \frac{800}{x^3}$

To find the critical point set C'(x)=0

$\therefore 0= - \frac{400}{x^2}+\frac53$

$\Rightarrow \frac{400}{x^2}=\frac{5}{3}$

$\Rightarrow x^2 =\frac{400\times 3}{5}$

$\Rightarrow x=\sqrt{240}$

$\Rightarrow x=15.49 \approx15.5$

Therefore

$\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0$

Therefore at x= 15.5 , C(x) is minimum.

Putting the value of x in $y=\frac{200}{x}$ we get

$\therefore y=\frac{200}{15.5}$

=12.9

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

6 0

3 years ago

You are trying to determine the half-life of a new radioactive element you have isolated. You start with 1 gram, and 5 days late

Oliga [24]

Half-life = elapsed time * log(2) / log (bgng amt / ending amt)
half-life = (5 * 0.30103) / log (1/.6)
half-life = (5 * 0.30103) / log (1/.6)
half-life = 1.505 / 0.221848750 
half-life = 6.785 days
Location for formulas & calculator
http://www.1728.org/halflife.htm

5 0

3 years ago