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Marysya12 [62]
2 years ago
12

3 (without using trig table) evaluate It tan 60 x tan 30 divided by Tan 60 + tan 30

Mathematics
1 answer:
Marizza181 [45]2 years ago
8 0

Answer:

1.25332

Step-by-step explanation:

tan(30 times  tan(60  )  \div  \tan(60 \div + 30)

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Sarah earned a grade of 80% on the math exam that had 75 problems. How many correct problems did Sara answer?​
xenn [34]

Answer:

60

Step-by-step explanation:

80/100 times 75 and you will get 60

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3 years ago
Solve for x please and thank uou
nata0808 [166]

Answer:

The question is not asked correctly

Step-by-step explanation:

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3 years ago
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(3x^2a - 4y^a z^3a) ^2 ...?
shusha [124]
The answer is 9x⁴ᵃ - 24x²ᵃyᵃz³ᵃ + 16 y²ᵃz⁶ᵃ


(a - b)² = a² - 2ab + b²
(3x²ᵃ - 4yᵃz³ᵃ)² = (3x²ᵃ)² - 2 * 3x²ᵃ * 4yᵃz³ᵃ + (4yᵃz³ᵃ)² =
                         = 3²x²ᵃ*² - 2 * 3 * 4 x²ᵃ * yᵃz³ᵃ + 4²yᵃ*²z³ᵃ*² =
                         = 9x⁴ᵃ - 24x²ᵃyᵃz³ᵃ + 16 y²ᵃz⁶ᵃ
8 0
4 years ago
A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume t
sladkih [1.3K]

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

a)

\frac{dy}{dt}\ \alpha\  y(1-y)

\frac{dy}{dt}=ky(1-y)

where k is the constant of proportionality, dy/dt =  rate at which the rumor spreads

b)

\frac{dy}{dt}=ky(1-y)\\\frac{dy}{y(1-y)}=kdt\\\int\limits {\frac{dy}{y(1-y)}} \, =\int\limit {kdt}\\\int\limits {\frac{dy}{y}} +\int\limits {\frac{dy}{1-y}}  =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln(\frac{y}{1-y}) =kt+c\\taking \ exponential \ of\ both \ sides\\\frac{y}{1-y} =e^{kt+c}\\\frac{y}{1-y} =e^{kt}e^c\\let\ A=e^c\\\frac{y}{1-y} =Ae^{kt}\\y=(1-y)Ae^{kt}\\y=\frac{Ae^{kt}}{1+Ae^{kt}} \\at \ t=0,y=10\%\\0.1=\frac{Ae^{k*0}}{1+Ae^{k*0}} \\0.1=\frac{A}{1+A} \\A=\frac{1}{9} \\

y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days

5 0
3 years ago
Given the foci of (8,0) and the difference of the focal radii = 6, find the equation of the hyperbola
k0ka [10]

Given :

The foci of hyperbola are (8,0) and (-8,0) .

The difference of the focal radii = 6.

To Find :

The equation of the hyperbola.

Solution :

We know, distance between foci is given by :

2c = 8 - (-8)

c = 8

Also, difference between the foci or focal distance is given by :

2a = 6

a = 3

Now, we know for hyperbola :

b^2 +a^2 = c^2\\\\b^2={c^2-a^2}\\\\b^2 = 64-9\\\\b^2 = 55

General equation of hyperbola is :

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\\\\\dfrac{x^2}{9}+ \dfrac{y^2}{55}=1

Hence, this is the required solution.

3 0
3 years ago
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