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Natalija [7]
3 years ago
10

What is the length of the longest side of an isosceles triangle with leg lengths 2x+12 and 5x-9 and base length 6x-2?

Mathematics
1 answer:
Fiesta28 [93]3 years ago
3 0
Your longest length is going to be 40

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Need help ASAP will mark brainliest
nalin [4]

Answer:

  5x -7y = 21

Step-by-step explanation:

A sketch can convince you that BC is a transversal perpendicular to parallel lines AB and CD. The question asks for an equation for CD, so we just need to write the equation of a line through D that is parallel to AB.

One way to do this is to equate the slopes of the parallel lines:

  ∆y/∆x for AB = ∆y/∆x for CD

  (y2 -y1)/(x2 -x1) = (y -2)/(x -7) . . . . . . where (x1, y1) = A; (x2, y2) = B; (7, 2) = D

  (4 -(-1))/(1 -(-6)) = (y -2)/(x -7)

  5(x -7) = 7(y -2) . . . . . . . . . . . . . cross multiply

  5x -7y = 21 . . . . . . . . . . . . . . . . add 35 -7y, simplify

_____

Note that the graph shows the line CD is named "b", and its equation is shown at upper left. Multiplying that equation by -1 gives the one shown here.

6 0
3 years ago
Aiden refills three token machines in an arcade. He puts twice the number in machine A as in machine B, and in machine C , he pu
Sonja [21]
It is a possibility of being 16 or 5/6
7 0
3 years ago
Read 2 more answers
How do I solve all this
Cerrena [4.2K]
Number five is .45 repeating
3 0
3 years ago
A light shines from the top of a pole 50ft high. a ball is dropped from the same height from a point 30 ft away from the light.
belka [17]
The speed of the ball is 
ds/dt = 32t
At t =1/2 s
ds/dt = 16 ft/s
The distance from the ground
50 - 16(1/2)^2 = 46 ft
The triangles formed are similar
50/46 = (30 + x)/x
x = 345 ft

50 / (50 - s) = (30 + x)/x
Taking the derivative and substituing
ds/dt = 16
and
Solve for dx/dt
7 0
3 years ago
Which equation is best represented by the graph above (x+1)(x-3)(x+2). please explain why
Nat2105 [25]
<h3>Answer:</h3>

y = (x +2)(x -1)(x -3) . . . . or . . . . y = x³ -2x² -5x +6

<h3>Step-by-step explanation:</h3>

The graph shows y=0 at x=-2, x=1, and x=3. These are called the "zeros" or "roots" of the function, because the value of the function is zero there.

When "a" is a zero of a polynomial function, (x -a) is a factor. This means the factors of the graphed function are (x -(-2)), (x -1) and (x -3). The function can be written as the product of these factors:

... y = (x +2)(x -1)(x -3) . . . . . the equation represented by the graph

Or, the product can be multiplied out

... y = (x +2)(x² -4x +3)

... y = x³ -2x² -5x +6 . . . . . the equation represented by the graph

7 0
3 years ago
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