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Tatiana [17]
2 years ago
8

PLEASE HELP !!!!!!!! prove that segment DE is parallel to segment AB.

Mathematics
2 answers:
RideAnS [48]2 years ago
3 0
Angle A is equal to Angle D because they are opposite interior angles I believe. Same thing goes for angles B and E.
GaryK [48]2 years ago
3 0

Slope of AB = (0 - 0)/(x2 - 0) = 0

Slope of DE = (y1/2 - y1/2) / ((x1 + x2)/2 - (x1/2)) = 0

Both lines has same slopes and equal zero.

Two lines are parallel if their slopes are equal. So AB ║ DE

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Simplify (12b*7) / (4b*5)
Alla [95]
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ First, just divide the numbers:

12/4 = 3

When dividing values with exponents, you have to subtract the exponents:

7 - 5 = 2

Therefore, your answer would be option B. 3b^2

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5 0
2 years ago
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Helppp meeee please
nekit [7.7K]

Answer:

The Equation for lyft is:

4.50x

The Equation for uber is

2.50x+5

Step-by-step explanation:

If you want to determine which one is better then you can just plug in the distance in place of x. For example if it asked you which one was better for 5 miles, uber or lyft you could plug 5 into the equation.

It would look like this

Lyft: 4.50 (5)= 22.5

Uber

2.50 (5) +5= 17.5

If the distance were 5 miles uber would be the better option.

5 0
1 year ago
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Why dose a triangle form 180 degress
Lelu [443]
All Triangles are equal to 180° on a plane surface no triangle is equal to 180° on a curved surface.
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3 years ago
Let g(x)=Intragal from 0 to x f(t) dt, where r is the function whos graph is shown.
leonid [27]

If

\displaystyle g(x) = \int_0^x f(t) \, dt

then g(x) gives the signed area under f(x) over a given interval starting at 0.

In particular,

\displaystyle g(0) = \int_0^0 f(t) \, dt = 0

since the integral of any function over a single point is zero;

\displaystyle g(4) = \int_0^4 f(t) \, dt = 8

since the area under f(x) over the interval [0, 4] is a right triangle with length and height 4, hence area 1/2 • 4 • 4 = 8;

\displaystyle g(8) = \int_0^8 f(t) \, dt = 0

since the area over [4, 8] is the same as the area over [0, 4], but on the opposite side of the t-axis;

\displaystyle g(12) = \int_0^{12} f(t) \, dt = -8

since the area over [8, 12] is the same as over [4, 8], but doesn't get canceled;

\displaystyle g(16) = \int_0^{16} f(t) \, dt = 0

since the area over [12, 16] is the same as over [0, 4], and all together these four triangle areas cancel to zero;

\displaystyle g(20) = \int_0^{20} f(t) \, dt = 24

since the area over [16, 20] is a trapezoid with "bases" 4 and 8, and "height" 4, hence area (4 + 8)/2 • 4 = 24;

\displaystyle g(24) = \int_0^{24} f(t) \, dt = 64

since the area over [20, 24] is yet another trapezoid, but with bases 8 and 12, and height 4, hence area (8 + 12)/2 • 4 = 40, which we add to the previous area.

5 0
2 years ago
The graph shows a system consisting of a linear equation and a quadratic equation.
Lorico [155]

Answer:

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Step-by-step explanation:

The points where the two curves intersect are the solutions to the system of equations:

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3 0
3 years ago
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