Help me please.

Please it URGENT i am being timed on a test

elena-14-01-66 [18.8K]

Answer:

angle 1 = angle 7 = angle 9 = angle 15 and angle 3 = angle 5 = angle 11 = angle 13

Also,

angle 2 = angle 8 = angle 10 = angle 16 and angle 4 = angle 6 = angle 12 = angle 14

Step-by-step explanation:

Let suppose a and b are two parallel lines, and c and d be the transverse lines.

The pairs of corresponding angles for a non-vertical line c are:

2-10, 7-15, 1-9, 8-16.

The pairs of corresponding angles for a non-vertical line d are:

4-12, 5-13, 3-11, 6-14.

Angles are formed at the intersection between the transverse lines and parallel lines. As the vertical angle is the opposite of the intersection. So, vertical angles are equal.

The pairs of vertical angles for a non-vertical line c are 1-7, 2-8, 9-15, 10-16

The pairs of vertical angles for a non-vertical line v are 3-5, 4-6, 11-13, 12-14

For transverse line c:

As the vertical angles are equal so the angles 1 = 7 = 9 = 15

As the corresponding angles are equal, so the angle ∠ 10 = ∠ 2 and the opposite of ∠2 is ∠8 which is vertical to ∠8 and vertical angles are also equal.

So, the angles 10 = 16 = 2 = 8 are also equal.

For transverse line d:

As the vertical angles are equal so the angles 3 = 5 = 11 = 13

As the corresponding angles are equal, so the angle ∠12 = ∠4 and the opposite of ∠4 is ∠6 which is vertical to ∠6 and vertical angles are also equal.

So, the angles 12 = 14 = 4 = 6 are also equal.

<u>LETS TAKE AN EXAMPLE</u>:

Let suppose the angle 1 is 110°

Since, vertical angle are equal. So, angle 7 will be also 110°.

Hence, angle 1 = 110° = angle 7

We also know that corresponding angles are equivalent. So,

angle 9 would also be equal to 110°, and as the opposite of angle 9 would also be equal to angle 9. So, the angle 9 and angle 15 have same angles.

So, angle 9 = 110° = angle 15

Hence, angle 1 = angle 7 = angle 9 = angle 15 will all be equal to 110°.

As, the angle 2 is the supplementary to this angle 1.

As the angle 1 is 110, so the angle 2 will be 70° as they are supplementary to each other.

And if angle 2 is 70°, then its opposite angle will be equal to 8. As angle 2 and 8 are equal.

So, angle 2 = 70° = angle 8

As the corresponding angles are equal, so the corresponding of 2 will be the angle 10 which will be also equal to 70°, and the opposite of angle 10 is angle 16, which is again equal to angle 10.

So, angle 10 = 70° = angle 16

Hence: angle 2 = angle 8 = angle 10 = angle 16 will be equal to 70°.

Since the transverse line d is parallel to c. So,

angle 1 = angle 7 = angle 9 = angle 15 which are equal to 110° will have same angles as of angle 3 = angle 5 = angle 11 = angle 13

So, angle 3 = angle 5 = angle 11 = angle 13 will also be equal to 110°

And,

angle 2 = angle 8 = angle 10 = angle 16 will which are equal to 70° will have same angles as of angle 4 = angle 6 = angle 12 = angle 14

So, angle 4 = angle 6 = angle 12 = angle 14 will be equal to 70°

Keywords: line, parallel lines, transverse line, angles

Learn More about simplifying radical expression from brainly.com/question/13209798

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Step-by-step explanation:

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Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

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Margaret has a monthly clothes budget of $50. She maps the amount of money she spends each month to the number of items of cloth

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Margaret has a monthly clothes budget of $50. This implies that she can spend a maximum of $50 on clothes and therefore the maximum value of the domain is 50 to or it can also be said that the domain is all values of domain from 0 to 50.

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