2x-6y= 12 3y=x+6 how many solutions does this problem have? No solution, one solution, or infinite solutions

Evaluate this please

AleksandrR [38]

This would give you 0.027 which equals 1/36

4 0

3 years ago

The scale below represents which equations or inequality?

77julia77 [94]

Answer:

c

Step-by-step explanation:

the scale is balanced so its equal

4 0

4 years ago

A town has 5000 people in year t = 0. Calculate how long it takes for the population P to double once, twice, and three times, a

UNO [17]

Answer:

Step-by-step explanation:

At the time t = 0, population of the town = 5000

Rate of population increase = 500 per year

Therefore, the equation that will represent the population will be

$P_{t}=P_{0}+500t$

Where $P_{t}$ = Population after t years

$P_{0}$ = Initial population

t = Time in years

a). For double once the population will be 500×2 = 10000

By plugging in the values in the equation,

10000 = 5000 + 500t

500t = 10000 - 5000

500t = 5000

t = $\frac{5000}{500}$

t = 10 years

For Double twice,

Population will be = 10000×2 = 20000

Now we plug in the values in the equation again

20000 = 5000 + 500t

500t = 20000 - 5000

500t = 15000

t = $\frac{15000}{500}$

t = 30 years

For double thrice,

Population of the town = 20000×2 = 40000

Now we plug in the values in the equation,

40000 = 5000 + 500t

500t = 40000 - 5000

500t = 35000

t = $\frac{35000}{500}$

t = 70 years

b). If the population growth is 5%.

Then the growth will be exponential represented by

$T_{n}=T_{0}(1+\frac{r}{100})^{t}$

$T_{n}$ = Population after t years

$T_{0}$ = Initial population

t = time in years

For double once,

Population after t years = 10000

$10000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{10000}{5000}$

$(1.05)^{t}=2$

Take log on both the sides

$log(1.05)^{t}=log2$

tlog(1.05) = log2

t = $\frac{log2}{log1.05}$

t = 14.20 years

For double twice,

Population after t years = 20000

$20000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{20000}{5000}$

$(1.05)^{t}=4$

Take log on both the sides

$log(1.05)^{t}=log4$

tlog(1.05) = log4

t = $\frac{log4}{log1.05}$

t = 28.413 years

For double thrice

Population after t years = 40000

$40000=5000(1+\frac{5}{100})^{t}$

$(1.05)^{t}=\frac{40000}{5000}$

$(1.05)^{t}=8$

Take log on both the sides

$log(1.05)^{t}=log8$

tlog(1.05) = log8

t = $\frac{log8}{log1.05}$

t = 42.620 years

4 0

3 years ago

What is the slope-intercept equation for the following line?

Lyrx [107]

Answer:

y = 2x + 7

Step-by-step explanation:

ALgebra is my best subject! Have a good day!

7 0

3 years ago

So, the sides you are going to be multiplying are opposite and adjacent. The trigonometric function is tangent. So it'll be Tan0

nadya68 [22]

Answer:

54°

Step-by-step explanation:

Multiplication is not involved.

The trig relation that is concerned with the opposite and adjacent sides of an angle in a right triangle is the tangent relation:

Tan = Opposite/Adjacent

tan(θ) = 41/30

The value of the angle is found using the inverse tangent function:

θ = arctan(41/30) ≈ 53.807°

The angle of elevation is about 54°.

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<em>Additional comment</em>

The mnemonic SOH CAH TOA is often used as a reminder of the relationship between right triangle sides and trig functions. The other two relations are ...

Sin = Opposite/Hypotenuse

Cos = Adjacent/Hypotenuse

For finding angles, the inverse trig relations are used. These can be called "arcsine," "arccosine," and "arctangent." They are often shown in functional form as (respectively) sin⁻¹, cos⁻¹, and tan⁻¹. On a calculator, they are often "2nd" or "Shift" or "Alt" functions of the sin, cos, and tan keys.

8 0

3 years ago