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statuscvo [17]
3 years ago
13

A 8.5 B 8.1 C 12.5 D 12.9

Mathematics
1 answer:
elena55 [62]3 years ago
4 0
Cosine again. Yay.

Put 36 into a fancy calculator with one of those cos() things. You get about 0.81. 

So that's the ratio between x and 10.

So basically x/10 = 0.81, or 0.81 * 10 = x.

8.1 = x
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What is the distance between the points P(-5, 4), Q(7, -5)
Nikolay [14]

Answer:

15

Step-by-step explanation:

Distance between two points is = √(x2 - x1)² + (y2 - y1)²

x1 = -5, y1 = 4; x2 = 7, y2 = -5

|PQ| = √(7 - (-5))² + (-5 - 4)²

|PQ| = √(7 + 5)² + (-9)²

|PQ| = √(12)² + 81

|PQ| = √144 + 81

|PQ| = √225

|PQ| = 15

4 0
3 years ago
¿cuanto pesa cada uno de nuestros 3 personajes?
sergij07 [2.7K]

Resolviendo el sistema de ecuaciones veremos que:

  • niña = 23kg
  • niño = 28kg
  • perro = 18kg.

<h3>¿Como resolver el sistema de ecuaciones?</h3>

Aqui tenemos el sistema de ecuaciones:

Niña + niño = 51kg

Niño + perro = 46 kg

Niña + perro = 41kg

Para resolver esto, lo primero que debemos hacer es aislar una variable en una de las ecuaciones, por ejemplo, podriamos aislar "perro" en la tercera:

perro = 41kg - niña

Ahora reemplazamos eso en la segunda para obtener:

niño + (41kg - niña) = 46kg

niño - niña = 46kg - 41kg = 5kg

niño = niña + 5kg

Ahora logramos obtener la variable "niño" en terminos de la variable "niña". Podemos reemplazar esto en la primera ecuacion del sistema.

niña + niño = 51kg

niña + (niña + 5kg) = 51kg

2*niña = 51kg - 5kg = 46kg

niña = 46kg/2 = 23kg.

Ahora que sabemos esto, usamos las otras ecuaciones para encontrar el peso del niño y el perro:

niño = niña + 5kg = 23kg + 5kg = 28kg

perro = 41kg - niña = 41kg - 23kg = 18kg.

Sí quieres aprender más sobre sistemas de ecuaciones, puedes leer:

brainly.com/question/17174746

6 0
2 years ago
Find the x-coordinate of Q' after Q(-2.5) was reflected over the line y = -x + 4.
kenny6666 [7]

Answer:

Q' would be (5, 6)

7 0
3 years ago
This is a picture of a cube and the net for the cube. What is the surface area of the cube? 78in 169in 507in or 1,014in
Stolb23 [73]

Answer:

1014 cm^2.

Step-by-step explanation:

There are 6 square faces on the cube and each face has area 13^2 = 169 cm^2.

The total area is 169 * 6 = 1.014 cm^2.

5 0
3 years ago
Read 2 more answers
The population of mosquitoes in a certain area increases at a rate proportional to the current pop-ulation, and in the absence o
ollegr [7]

Answer:

The population of mosquitoes in the area at any time <em>t</em> is:

P(t)=201977.31-1977.31\times 2^{t}

Step-by-step explanation:

The rate of growth of mosquitoes can be expressed as:

\frac{dP}{dt}=kP

\frac{dP}{P}=k\ dt

Integrate the above expression as follows:

\int {\frac{dP}{P}} \, =\int {k\ dt} \, \\\ln|P|=kt+c\\e^{\ln|P|}=e^{kt+c}\\P=Ce^{kt}

\Rightarrow P=P_{0}e^{kt}

It is provided that the population doubles every day.

Compute the value of <em>k</em> as follows:

2=1\times e^{k\times1}\\2=e^{k}\\k=\ln (2)

It is also provided that every day 20,000 mosquitoes are eaten.

The rate of growth per week can be expressed as:

\frac{dP}{dt}=\ln(2)P-14000\\\frac{dP}{dt}-\ln(2)P=14000

The integrating factor for this is:

e^{\int {\ln(2)dP}}=e^{\ln(2)\int {dt}}=e^{\ln(2)t}

Then,

P(t)\ e^{-\ln(2)t}=\int {e^{-\ln(2)t}}-14000\, dt\\=-14000\int {e^{-\ln(2)t}}\, dt\\=-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C\\P(t)=(e^{-\ln(2)t})\times [-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C]\\=\frac{14000}{\ln(2)}+Ce^{-\ln(2)t}

The initial population is 200,000.

Compute the value of <em>C</em> as follows:

P(t)=\frac{140000}{\ln(2)}+Ce^{-\ln(2)t}\\200000=\frac{14000}{\ln(2)}+Ce^{-\ln(2)(0)}\\C=200000-\frac{140000}{\ln(2)}\\C=-1977.31

Now substitute <em>C</em> in P (t),

P(t)=\frac{140000}{\ln(2)}+Ce^{\ln(2)t}\\P(t)=201977.31-1977.31\times 2^{t}

6 0
2 years ago
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