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statuscvo [17]
2 years ago
11

What is the value of Q?

Mathematics
1 answer:
butalik [34]2 years ago
4 0

Answer:

52 degrees

Step-by-step explanation:

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b -1

Step-by-step explanation:

it's the line that the curve ends on and is linear on

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In 3 hours, a car travels 120 miles. What is the speed of the car
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Si un celular Cuesta 4500 y tiene un descuento de 76%cuamto vq a pagar por el teléfono
USPshnik [31]

De acuerdo con las definiciones de precio a pagar, porcentaje de descuento y porcentaje remanente, el precio a pagar por el celular es $ 1080.

<h3>¿Cuánto cuesta un celular con descuento? </h3>

En esta situación una persona recibe un descuento por la compra de su celular, el precio a pagar es el producto de multiplicar el coste <em>sin descuento</em> y el porcentaje remanente en forma fraccionaria. El porcentaje remanente se calcula al restar el descuento del cien por ciento. A continuación, procedemos al cálculo requerido:

x = 4500 × (1 - 76/100)

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De acuerdo con las definiciones de precio a pagar, porcentaje de descuento y porcentaje remanente, el precio a pagar por el celular es $ 1080.

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1 year ago
I really need help or I’m gonna fail the semester
Tanzania [10]
Rational Number, Real Number
3 0
2 years ago
If N is the population of the colony and t is the time in​ days, express N as a function of t. Consider Upper N 0 is the origina
lesya [120]

Answer:

(a)N(t)=Noe^{kt}

(b)5,832 Mosquitoes

(c)5 days

Step-by-step explanation:

(a)Given an original amount N_o at t=0. The population of the colony with a growth rate k \neq 0, where k is a constant is given as:

N(t)=Noe^{kt}

(b)If N_o=1000 and the population after 1 day, N(1)=1800

Then, from our model:

N(1)=1800

1800=1000e^{k}\\$Divide both sides by 1000\\e^{k}=1.8\\$Take the natural logarithm of both sides\\k=ln(1.8)

Therefore, our model is:

N(t)=1000e^{t*ln(1.8)}\\N(t)=1000\cdot1.8^t

In 3 days time

N(3)=1000\cdot1.8^3=5832

The population of mosquitoes in 3 days time will be approximately 5832.

(c)If the population N(t)=20,000,we want to determine how many days it takes to attain that value.

From our model

N(t)=1000\cdot1.8^t\\20000=1000\cdot1.8^t\\$Divide both sides by 1000\\20=1.8^t\\$Convert to logarithm form\\Log_{1.8}20=t\\\frac{Log 20}{Log 1.8}=t\\ t=5.097\approx 5\; days

In approximately 5 days, the population of mosquitoes will be 20,000.

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2 years ago
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