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erik [133]
2 years ago
13

A store had 4 boxes of video Games.How many days would it take to sell the games if each day they sold one fifth

Mathematics
2 answers:
Korvikt [17]2 years ago
8 0
The answer would be 20 days.
Mekhanik [1.2K]2 years ago
5 0
It would tale 20 days 
 
5*4=20 you take away the fraction part of one fifth:)) 

hope it helped!!!!
You might be interested in
The triangles in the diagram are congruent. If mF = 40°, mA = 80°, and mG = 60°, what is mB?
maria [59]

Answer:

The answer is 40 degrees.

Step-by-step explanation:

Since the two triangles are congruent, which means they are the same shape and size, they share the same angle measures. Since angle B is the same as angle F, it equals 40 degrees.


I hope this helped! :-)

5 0
2 years ago
Read 2 more answers
Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
2 years ago
La fuerza necesaria para evitar que un auto derrape en una curva varía inversamente al radio de la curva y conjuntamente con el
Vika [28.1K]

Answer:

768 libras de fuerza

Step-by-step explanation:

Tenemos que encontrar la ecuación que los relacione.

F = Fuerza necesaria para evitar que el automóvil patine

r = radio de la curva

w = peso del coche

s = velocidad de los coches

En la pregunta se nos dice:

La fuerza requerida para evitar que un automóvil patine alrededor de una curva varía inversamente con el radio de la curva.

F ∝ 1 / r

Y luego con el peso del auto

F ∝ w

Y el cuadrado de la velocidad del coche

F ∝ s²

Combinando las tres variaciones juntas,

F ∝ 1 / r ∝ w ∝ s²

k = constante de proporcionalidad, por tanto:

F = k × w × s² / r

F = kws² / r

Paso 1

Encuentra k

En la pregunta, se nos dice:

Suponga que 400 libras de fuerza evitan que un automóvil de 1600 libras patine alrededor de una curva con un radio de 800 si viaja a 50 mph.

F = 400 libras

w = 1600 libras

r = 800

s = 50 mph

Tenga en cuenta que desde el

F = kws² / r

400 = k × 1600 × 50² / 800

400 = k × 5000

k = 400/5000

k = 2/25

Paso 2

¿Cuánta fuerza evitaría que el mismo automóvil patinara en una curva con un radio de 600 si viaja a 60 mph?

F = ?? libras

w = ya que es el mismo carro = 1600 libras

r = 600

s = 60 mph

F = kws² / r

k = 2/25

F = 2/25 × 1600 × 60² / 600

F = 768 libras

Por lo tanto, la cantidad de fuerza que evitaría que el mismo automóvil patine en una curva con un radio de 600 si viaja a 60 mph es de 768 libras.

7 0
2 years ago
A right cylinder has a radius of 2 units and height of 5 units.What is the volume of the cylinder? Round to the nearest tenth.
UkoKoshka [18]
V=hpir^2
r=2
h=5
pi≈3.141592

v=5*3.141592*2^2
v=5*3.141592*4
v=20*3.141592
v=62.83185307179586476925286766559
round to tenth
62.8 cubic units
8 0
2 years ago
Read 2 more answers
A trapezoid has a base of 4.5 inches, a height of 6 inches, and an area of 21 square inches. The equation below can be used to d
Troyanec [42]

Answer:

b_2 =  -2.5

Step-by-step explanation:

Given

\frac{1}{2}(4.5 + b_2) * 6 = 21

Required

Determine the value of T

\frac{1}{2}(4.5 + b_2) * 6 = 21

Multiply both sides by 2

2 * \frac{1}{2}(4.5 + b_2) * 6 = 21 * 2

(4.5 + b_2) * 6 = 42

Divide through by 6

4.5 + b_2 = 42/6``

4.5 + b_2 = 7

b_2 =  7 - 4.5

b_2 =  -2.5

6 0
2 years ago
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