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andreev551 [17]
1 year ago
13

In the diagram, the measure of angle 2 is 126°, the measure of angle 4 is (7x)°, and the measure of angle 5 is (4x + 4)°. A tran

sversal intersects 2 lines to form 8 angles. Clockwise from the top left, the angles are 1, 2, 3, 4; 5, 6, 7, 8. What is the measure of angle 7, to the nearest degree?
Mathematics
2 answers:
tigry1 [53]1 year ago
6 0

Answer:

  1. 7 is (14x)°

Step-by-step explanation:

In the diagram, the measure of angle 2 is 126°, the measure of angle 4 is (7x)°, and the measure of angle 5 is (4x + 4)°. A transversal intersects 2 lines to form 8 angles. Clockwise from the top left, the angles are 1, 2, 3, 4; 5, 6, 7, 8. What is the measure of angle 7, to the nearest degree? The nearest degree measure of angle 7 is (14x)°

Lena [83]1 year ago
5 0

Answer:

The correct option is C

C) 76°

Step-by-step explanation:

The question is incomplete because the figure is not given. I have attached the figure of the same question below. Consult it for better understanding.

We can see in the figure that <2 and <4 are opposite angles, therefore these are equal.

<2 = <4

126 = 7x

x = 18

Similarly we can also see that <5 and <7 are also opposite angles, therefore they are equal too.

<7 = <5

<7 = 4x+4

Substitute x=18

<7 = 4(18) + 4

<7 = 76°

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2 years ago
A group of neighborhood children have a lemonade stand. They spend $5 on supplies. The function graphed shows the earnings they
Elanso [62]

Hey!


Your answer would be:


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7 0
1 year ago
If a is an arbitrary nonzero constant, what happens to a/b as b approaches 0
Stolb23 [73]

It depends on how b approaches 0

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7 0
2 years ago
Suppose the clean water of a stream flows into Lake Alpha, then into Lake Beta, and then further downstream. The in and out flow
Gala2k [10]

Answer:

a) dx / dt = - x / 800

b) x = 500*e^(-0.00125*t)

c) dy/dt = x / 800 - y / 200

d) y(t) = 0.625*e^(-0.00125*t)*( 1  - e^(-4*t) )

Step-by-step explanation:

Given:

- Out-flow water after crash from Lake Alpha = 500 liters/h

- Inflow water after crash into lake beta = 500 liters/h

- Initial amount of Kool-Aid in lake Alpha is = 500 kg

- Initial amount of water in Lake Alpha is = 400,000 L

- Initial amount of water in Lake Beta is = 100,000 L

Find:

a) let x be the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash. find a formula for the rate of change in the amount of Kool-Aid, dx/dt, in terms of the amount of Kool-Aid in the lake x:

b) find a formula for the amount of Kook-Aid in kilograms, in Lake Alpha t hours after the crash

c) Let y be the amount of Kool-Aid, in kilograms, in Lake Beta t hours after the crash. Find a formula for the rate of change in the amount of Kool-Aid, dy/dt, in terms of the amounts x,y.

d) Find a formula for the amount of Kool-Aid in Lake Beta t hours after the crash.

Solution:

- We will investigate Lake Alpha first. The rate of flow in after crash in lake alpha is zero. The flow out can be determined:

                              dx / dt = concentration*flow

                              dx / dt = - ( x / 400,000)*( 500 L / hr )

                              dx / dt = - x / 800

- Now we will solve the differential Eq formed:

Separate variables:

                              dx / x = -dt / 800

Integrate:

                             Ln | x | = - t / 800 + C

- We know that at t = 0, truck crashed hence, x(0) = 500.

                             Ln | 500 | = - 0 / 800 + C

                                  C = Ln | 500 |

- The solution to the differential equation is:

                             Ln | x | = -t/800 + Ln | 500 |

                                x = 500*e^(-0.00125*t)

- Now for Lake Beta. We will consider the rate of flow in which is equivalent to rate of flow out of Lake Alpha. We can set up the ODE as:

                  conc. Flow in = x / 800

                  conc. Flow out = (y / 100,000)*( 500 L / hr ) = y / 200

                  dy/dt = con.Flow_in - conc.Flow_out

                  dy/dt = x / 800 - y / 200

- Now replace x with the solution of ODE for Lake Alpha:

                  dy/dt = 500*e^(-0.00125*t)/ 800 - y / 200

                  dy/dt = 0.625*e^(-0.00125*t)- y / 200

- Express the form:

                               y' + P(t)*y = Q(t)

                      y' + 0.005*y = 0.625*e^(-0.00125*t)

- Find the integrating factor:

                     u(t) = e^(P(t)) = e^(0.005*t)

- Use the form:

                    ( u(t) . y(t) )' = u(t) . Q(t)

- Plug in the terms:

                     e^(0.005*t) * y(t) = 0.625*e^(0.00375*t) + C

                               y(t) = 0.625*e^(-0.00125*t) + C*e^(-0.005*t)

- Initial conditions are: t = 0, y = 0:

                              0 = 0.625 + C

                              C = - 0.625

Hence,

                              y(t) = 0.625*( e^(-0.00125*t)  - e^(-0.005*t) )

                             y(t) = 0.625*e^(-0.00125*t)*( 1  - e^(-4*t) )

6 0
2 years ago
En la escuela Francisco I. Madero de ciudad Delicias, llevaron a cabo el festejo para
REY [17]

Answer:

The number of unsold cakes was 2

Step-by-step explanation:

<u><em>The question in English is</em></u>

In the school Francisco I. Madero of Ciudad Delicias, the celebration was held for  commemorate the arrival of spring, after the parade the stalls were set up  of the kermesse. The first grade group bought 8 cakes and sold 3/4 of the total.

How much of the cake was not sold?

Let

x ----> number of cakes sold

y ----> number of cakes that didn't sell

we know that

The first grade group bought 8 cakes

so

x+y=8 -----> equation A

The first grade group sold 3/4 of the total.

so

x=\frac{3}{4}(x+y) ---> equation B

substitute equation A in equation B

x=\frac{3}{4}(8)=6

Find the value of y

6+y=8

y=2

therefore

The number of unsold cakes was 2

4 0
1 year ago
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