All categories

4 years ago

What is that as a single power

Mathematics

2 answers:

denis-greek [22]4 years ago

8 0

Answer:

1.2

Step-by-step explanation:

3 over 3 and 2 over 3 is 1.2

Damm [24]4 years ago

7 0

Answer:

(3/2)power 3 = 27/8 =3.375

You might be interested in

For a filed trip the school bought 47 sandwiches for $40.60 each and 39 bags of chips for $1.25 each. How much did the school sp

lyudmila [28]

Answer:

$1,956.95

Step-by-step explanation:

The school bought 47 sandwiches for $40.60

= 47 × 40.60

= 1,908.2

They also bought 39 bags of chips for $1.25

= 1.25×39

= 48.75

Therefore the total amount spent can be calculated as follows

= 48.75 + 1,908.2

= 1,956.95

Hence the total amount spent by the school is $1,956.95

6 0

3 years ago

Finding Distance in the Coordinate Plane!

Aleonysh [2.5K]

It is B. the differences are added together

7 0

4 years ago

A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago

Add three data values to the following data set so the mean increases by 10 and the median does not change. {42, 37, 32, 29, 20}

aksik [14]

Sum of the numbers in the set: 42+37+32+29+20 =160

Current mean: 160/5 = 32

Median = the valvue of the middle = 32.

New mean: 32+10= 42.

Sum of numbers in the new set = 42*8 = 336

Difference: 336 - 160 = 176

I want to include 32, so that the new median stays in 32.

So the other two numbers must add 176 - 32 = 144

I will use a smaller number than 32 and the other greater (again in order to keep the same median.

I will choose 28 and 144 - 28 = 116.

So my three new numbers are 28, 32 and 116 and the new set is {116, 42, 37, 32, 32, 29,28, 20}

Checking:

Sum of the terms: 116+42+37+32+32+29+28+20 = 336

Mean = 336 / 8 = 42, which is 10 more than the original mean.

And the new median is (32+32)/2 = 32. The same of the original set.

8 0

3 years ago

Choose the point on the terminal side of -210°. Please only answer this question if you are willing to show work so I can learn

N76 [4]

X=rcos∅=r cos (-210)=r cos 210=rcos (180+30)=-r cos 30
$x=-r* \frac{sqrt 3}{2} = \frac{- sqrt 3r}{2} ,$
$y=r sin (-210)=-rsin 210=-rsin (180+30)=-r(-sin 30)$
1 $or y=r sin 30=r* \frac{ 1}{2} = \frac{r}{2}$

3 0

3 years ago