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2 years ago

Find the value of x and thank for answer

Mathematics

1 answer:

lakkis [162]2 years ago

3 0

The value for x is 102 degrees

First we can find the second value in the triangle by subtracting 141 from 180. This works because when added together these two angles come together to from a line (180 degrees) because this value is now known (39 degrees) we can subtract both from 180:

180-39-39= x = 102

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egoroff_w [7]

Answer:

Step-by-step explanation:

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3 years ago

Using a table of values, determine the solution to the equation below to the nearest fourth of a unit. 2^x=1-3^x

Greeley [361]

Answer:

Option (1)

Step-by-step explanation:

Given equation is,

$2^x=1-3^x$

To determine the solution of the equation we will substitute the values of 'x' given in the options,

Option (1)

For x = -0.75

$2^{-0.75}=1-3^{-0.75}$

0.59 = 1 - 0.44

0.59 = 0.56

Since, values on both the sides are approximately same.

Therefore, x = -0.75 will be the answer.

Option (2)

For x = -1.25

$2^{-1.25}=1-3^{-1.25}$

0.42 = 1 - 0.25

0.42 = 0.75

Which is not true.

Therefore, x = -1.25 is not the answer.

Option (3)

For x = 0.75

$2^{0.75}=1-3^{0.75}$

1.68 = 1 - 2.28

1.68 = -1.28

Which is not true.

Therefore, x = 0.75 is not the answer.

Option (4)

For x = 1.25

$2^{1.25}=1-3^{1.25}$

2.38 = 1 - 3.95

2.38 = -2.95

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3 years ago

Sauron [17]

Answer:

Step-by-step explanation:

"Four times a number" in symbols is "4n."

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3 years ago

What is the approximate volume of the cone?

Vlad [161]

Volume=1/3π<span>×</span>r²×h^(Applies for any "right cone")^
Therefore your answer would be
A.1206 cm³

8 0

3 years ago

Read 2 more answers

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

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3 years ago