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AlexFokin [52]
3 years ago
8

Two sides of a triangle have lengths 9 and 14 which inequalities describe the value that possible lengths for the third dide

Mathematics
1 answer:
lubasha [3.4K]3 years ago
3 0
| side1 - side2 | < side3 < ( side1 + side2 )

| 9 - 14 | < side3 < ( 9 + 14 )

5 < side3 < 23

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Marcelis takes a taxi to the airport. The taxi charges $2 per mile and an initial fee of
Marizza181 [45]

Answer:

13 miles

Step-by-step explanation:

27.50- 1.50(fee)=26.00 26/2(2$per mile)=13 miles

4 0
3 years ago
Can I have some help with this problem
Rufina [12.5K]
Corresponding measures have the same ratio.
.. (x -1)/4 = (2x +1)/10
.. 5(x -1) = 2(2x +1) . . . . . . multiply by 20
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.. x = 7 . . . . . . . . . . . . . . . add 5 -4x

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A store offers a 15% discount on all items in the store during a sale. All store employees get an additional 10% employee discou
vodomira [7]
The question tells us to find or to calculate how much would Chris pay for the coat during the sale. So to calculate the you must first analyze the formula that is given which is C=0.765+0.06(0.765x) while the variablec is the cost of an item and the X represents the price of the coat and the price of the coat on sale would be 81.09 
3 0
3 years ago
Please help me with this one
Rasek [7]

Answer:

6 (2)^ = 6

6 (2)^2 = 24

a = 24

b = 6

y-intercept: (0,6)

y-intercept =

Step-by-step explanation:

6(2)^x

6 (2)^0 = 6 (a number ^0 = 1, 6 times 1 = 6)

6 (2)^2 = 6 x 4 = 24

8 0
3 years ago
Suppose that receiving stations​ X, Y, and Z are located on a coordinate plane at the points ​(4​,5​), ​(-6​,-6​), and ​(-14​,2​
Lilit [14]

Answer:

  (-2, -3)

Step-by-step explanation:

A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.

_____

The simultaneous equations for the circles can be solved algebraically.

The epicenter is 10 units from X, so lies on the circle ...

  (x -4)^2 +(y -5)^2 = 10^2

  x^2 -8x +16 +y^2 -10y +25 = 100

  x^2 +y^2 -8x -10y = 59

__

The epicenter is 5 units from Y, so lies on the circle ...

  (x +6)^2 +(y +6)^2 = 5^2

  x^2 +12x +36 +y^2 +12y +36 = 25

  x^2 +y^2 +12x +12y = -47

__

The epicenter is 13 units from Z, so lies on the circle ...

  (x +14)^2 +(y -2)^2 = 13^2

  x^2 +28x +196 +y^2 -4y +4 = 169

  x^2 +y^2 +28x -4y = -31

__

Subtracting the second equation from each of the other two, we get ...

  (x^2 +y^2 -8x -10y) -(x^2 +y^2 +12x +12y) = (59) -(-47)

  -20x -22y = 106 . . . . eq1 -eq2

  (x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)

  16x -16y = 16 . . . . . . . .eq3 -eq2

These simultaneous linear equations can be solved a variety of ways. We might use substitution:

  x = y+1 . . . . . from eq3 -eq2 divided by 16

  10(y +1) +11y = -53 . . . . . from eq1 -eq2 divided by -2

  21y = -63 . . . . . . . . . . . . simplify, subtract 10

  y = -3

  x = y+1 = -2

The epicenter is located at (x, y) = (-2, -3).

8 0
3 years ago
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