All categories

3 years ago

I accidentally ordered two thousand boxes each containing fifty plastic hula dancers.

Mathematics

2 answers:

nadya68 [22]3 years ago

5 0

You would have 100000 hula dancers.

<h3>EXPLANATION</h3>

You would do 2000*50 to get 100000 hula dancers.

Hope this helps!

klemol [59]3 years ago

4 0

100,000 hula dancers

You might be interested in

Which two equations are true?(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)=6.6×10−10 (6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.

tiny-mole [99]

let me edit your question as:

Which two equations are true?

Eq1:

(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)

Eq2:

6.6×10−10(6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.7×103)

Eq3:

2.7×103(7.5×106)−(2.5×106)=5×100

Answer:

No one is true

Step-by-step explanation:

let's check each equation, if the values on both sides (left and right side) are equal then the equation is true otherwise false.

Using PEMDAS rule we are simplifying the equations as;

Eq1:

$(2*10-4)+(1.5*10-4)=3.5*10-4(3*10-5)+(2.2*10-5)\\(16)+(11)=35-4(25)+(17)\\27=35-100+17\\27=-48\\$

Eq2:

$6.6*10-10(6.3*10-1)-(2.1*10-1)=3*10-1(5.4*103)-(2.7*103)\\66-10(62)-(20)=30-1(556.2)-278.1\\66-620-20=30-556.2-278.1\\-574=-804.1$

Eq3:

$2.7*103(7.5*106)-(2.5*106)=5*100\\221089.5-265=500\\220824.5=500\\$

we observed that none of the equation has two same values on both sides thus none of the three equations is true.

Also, no value of Eq1, Eq2 or Eq3 are same thus none of the equation is true

8 0

3 years ago

Tamia is collecting boxes to help her friend move. She found a box that is 8 ft wide, 2 ft high, and 4 ft long how much space wo

Arada [10]

Answer: $64\ ft^3$

Step-by-step explanation:

Given

The dimension of the box is

$\text{width w=}8\ ft\\\text{height h=}2\ ft\\\text{length l=}4\ ft$

Space occupied by the box is equal to its volume i.e.

$\Rightarrow V=l\times b\times h\\\Rightarrow V=4\times 8\times 2\\\Rightarrow V=64\ ft^3$

The volume of the box is $64\ ft^3$ .

4 0

3 years ago

Prince Ivan rides gray wolf and a constant speed from kings castle to the magic apple garden in five hours. on the return trip t

borishaifa [10]

Answer:

The original speed of the grey wolf can be 9 and/or 48 km/hour

Step-by-step explanation:

There are two trips. One to King Castle (KC) to Magic Apple Garden (MAG), and then back (MAG - KC).

First trip: KC - MAG

It takes the same distance as MAG - KC so distance can be D for both.

Then it is stated that it takes 5 hours or 300mins for him to go from KC - MAG, and the speed is constant.

Let x be speed.

So the equation for this is:

D=5x

Second trip: MAG - KC

Here he goes 36 mi at constant speed (x), then picks up his speed by 3 km/hour (x+3).

For Time:

He reaches to KC, 15 mins earlier than KC - MAG, so he reaches there in 4.75 hours or in 275 min.

The problem splits into two as there are two different times, for the two parts of the journey or distance for MAG - KC.

Let T#1 = a, and T#2 = b.

Let D#2 = D-36

The equation for a:

a = 36/x --> This is he is at a constant speed and went for 36mi.

The equation for b:

b = (x+3)/D#2

These two equation have to be added as (a+b) = 4.75 --> 4.75 is the total time for MAG - KC.

So the equation is:

36/x + (5x-36)/x+3 = 4.75 --> Knowing (a+b) = 4.75, just put 4.75 as the equal, and add the fractions that equal (a) and (b).

Now Solve:

36/x + (5x-36)/x+3 = 4.75

0.25s^2 + 108 + 5s^2 - 36s = 4.75s*(s+3)

0.25s^2 - 14.25s + 108 = 0

(14.25 +- Square root (203.06 - 108))/0.5

(14.25 +- Square root (95.0625))/0.5

(14.25 +- 9.75)/0.5

Answer:

s = 9, s=48.

Therefore the original speed of the grey wolf can be 9 and/or 48 km/hour.

8 0

3 years ago

a parabola contains the points (0,0),(-1,-2) and (1,6). what is the equation of the parabola in standard form?

tensa zangetsu [6.8K]

Let the equation be:

y = ax^2 + bx + c.

Then, substitue the three points into the equation.

First point: 0 = a0^2 + b0 + c.

So c = 0.

Second point: -2 = a(-1)^2 + b(-1) + c.

So a - b + c = -2.

Third point: 6 = a*1^2 + b*1 + c.

So a + b + c = 6.

We know that c=0 already, so we substitute c=0 into the last two equations and we would get:

a - b = -2

a + b = 6

We add the two equations and we get:

2a = 4

a = 2

Then, we substitute a=2 into a-b=-2 and we get:

-b = -4

b = 4

Now we know a = 2, b = 4, and c = 0

Then, the equation of the parabola would be:

2x^2 + 4x

5 0

3 years ago

Determine the type and number of solutions of −4x^2 − 4x − 1 = 0

user100 [1]

The question is about types and number of soultions so you must use b^2-4ac which is the experssion under the root in the quadratic formula
so
if
b^2-4ac=0 then there is one real soultion
if
b^2-4ac<0 then there are two imaginary soultions that differ in the sign between them
if
b^2-4ac>0 then there are two real soultions
so a polynomial is written in the form
ax^2+bx+c=0
so a=-4
b=-4
c=-1
so b^2-4ac=16-16=0
so one real soultion is the answer you are looking
type:real
number:1

4 0

3 years ago