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3 years ago

a map shows a railroad crossing on a highway as shown below which of the numbered angles are supplementary

Mathematics

1 answer:

zimovet [89]3 years ago

4 0

Supplementary angles are angles that add up to 180 so it would probably be 2 and 3 because the highway is 180°

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<img src="https://tex.z-dn.net/?f=2%284%29%28%20%5Cfrac%7B1%7D%7B16%7D%20%29%20%3D%20" id="TexFormula1" title="2(4)( \frac{1}{16

Radda [10]

Answer:

0.5

Step-by-step explanation:

$2*4*\frac 1 {16}\\8 / 16\\0.5$

Hope this helped :)

7 0

3 years ago

Can someone explain how to graph this? I know how to graph it in the y= format. So can someone explain how to get these equation

artcher [175]

Answer:

hope this helps.

Step-by-step explanation:

You don't really need to get them in the y format.

make a table and plug some numbers in and graph those points

use a ruler to connect the dots.

y = -3 is a horizontal line

8 0

3 years ago

What is the solution to the equation?

Lisa [10]

$3q + 5 = -2q + 20$

add 2q to both sides.

$5q + 5 = 20$

subtract 5 to both sides.

$5q = 15$

divide 5 to both sides.

$q = \frac{15}{5}$

the answer is: $q = 3$

5 0

3 years ago

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.

7 0

3 years ago

If y=80,when x=32 find x when y=100

Bess [88]

Y X
80 32
100 x'

80x' = 100*32
80x' = 3200
x' = 3200/80
x' = 320/8
x' = 40

5 0

3 years ago

a map shows a railroad crossing on a highway as shown below which of the numbered angles are supplementary​

a map shows a railroad crossing on a highway as shown below which of the numbered angles are supplementary