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4 years ago

3.141 in expanded notation

Mathematics

2 answers:

Eva8 [605]4 years ago

8 0

3.141 in expanded notation is 3 + 0.1 + 0.04 + 0.001

Hope that helps :)

kari74 [83]4 years ago

6 0

3 + .1 + .04 + .001 is 3.141 in expanded notation.

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Determine the down payment for this vehicle: $28,900 truck at 24% down

WINSTONCH [101]

Based on the information given, the value of the down payment will be $6936.

<h3>Solving the percentage.</h3>

Based on the information given, the cost of the truck is $28900 and the down percentage given is 24%.

Therefore, the down payment that will be made will be:

= 24% × $28900

= 0.24 × $28900

= $6936

In conclusion, the correct option is $6936.

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brainly.com/question/24304697

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2 years ago

Craig collects coins. He collected a total of 225 coins. If 88% of coins he collected were foreign, how many other coins did he

agasfer [191]

Answer:

Step-by-step explanation:

In this case we associate in total value from 225 to 100%, and they tell us that foreign currencies represent 88%, therefore, in quantity this would be:

225 * 88/100

198

The rest of the coins would then refer to the other coins I collect, which would be:

225 - 198 = 27

That is, 27 coins are the remaining coins.

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3 years ago

Find the slope of the line that passes through (42, -74) and (70, -80).

PolarNik [594]

- 3/14 is your final answer.

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3 years ago

a person invested $6700 for one year, part at 8%, part at 10%, and the remainder at 12%. the total annual income from these inve

lukranit [14]

Answer:

The amount invested at 8% rate is $1,200

The amount invested at 10% rate is $2,000

The amount invested at 12% rate is $3,500

Step-by-step explanation:

step 1

Let

x-----> the amount invested at 8% rate

y-----> the amount invested at 10% rate

z-----> the amount invested at 12% rate

$z=(x+y)+300$ ----> equation A

$x+y+z=6,700$ ----> equation B

substitute equation A in equation B

$x+y+(x+y+300)=6,700$

$2x+2y=6,400$

$x+y=3,200$ -----> equation C

we know that

The simple interest formula is equal to

$I=P(rt)$

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

$t=1\ year\\ P=\$6,700\\ I=\$716$

substitute in the formula above

$716=x(0.08)+y(0.10)+z(0.12)$

substitute equation A

$716=x(0.08)+y(0.10)+(x+y+300)(0.12)$

$716=0.08x+0.10y+0.12x+0.12y+36$

$716=0.20x+0.22y+36$

$0.20x+0.22y=680$ -----> equation D

step 2

Solve the system of equations

x+y=3,200 -----> equation C

$0.20x+0.22y=680$ -----> equation D

Solve the system by graphing

The solution is the point (1,200,2,000)

see the attached figure

Find the value of z

$z=(x+y)+300$

$z=(1,200+2,000)+300=3.500$

therefore

The amount invested at 8% rate is $1,200

The amount invested at 10% rate is $2,000

The amount invested at 12% rate is $3,500

8 0

4 years ago

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

6 0

3 years ago