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lidiya [134]
3 years ago
12

Write the eqauation of a line that has the slope pf 6 passes through (-3,5)

Mathematics
1 answer:
Darina [25.2K]3 years ago
6 0
Y = mx + b
slope(m) = 6
(-3,5)...x = -3 and y = 5
now we sub and find b, the y int
5 = 6(-3) + b
5 = -18 + b
5 + 18 = b
23 = b

so ur equation is : y = 6x + 23
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WILL MAKR BRAINLIEST Jack jogs and rides his bike for aWrite a pair of linear equations to show the relationship between the num
PilotLPTM [1.2K]

Answer:

The pair of linear equations is

x+y = 75

y+15 = x

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y = number of minutes Jack bikes

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x+y = 75

He bikes for 15 minutes longer than he jogs

y+15 = x

The pair of linear equations is

x+y = 75

y+15 = x

8 0
3 years ago
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What is the midpoint between (4,12) and (8,2)
Maurinko [17]
The distance between the x values is 4 and the distance between the y values is 10...

divide the distance between the x values and the y values
add 2 to the 4 on x
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8 0
3 years ago
Represents a exponential form
lidiya [134]
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6 0
4 years ago
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Answer it and show your work. Which letter choice is it.
ivanzaharov [21]

Answer:

b = 55 degrees

Step-by-step explanation:

The angles 30, a, b and 45 must sum up to 180 degrees

Subtracting (30 + 45) from both sides leaves us with a + b = 105 degrees.

But b = a + 5.  Substituting a + 5 in the equation above yields

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7 0
3 years ago
Answer for a lot of points!
earnstyle [38]

Given :

  • ZC = 90°

  • CD is the altitude to AB.

  • \angleA = 65°.

To find :

  • the angles in △CBD and △CAD if m∠A = 65°

Solution :

In Right angle △ABC,

we have,

=> ACB = 90°

=> \angleCAB = 65°.

So,

=> \angleACB + \angleCAB+\angleZCBA = 180° (By angle sum Property.)

=> 90° + 65° + \angleCBA = 180°

=> 155° +\angleCBA = 180°

=> \angleCBA = 180° - 155°

=> \angleCBA = 25°.

In △CDB,

=> CD is the altitude to AB.

So,

=> \angle CDB = 90°

=> \angleCBD = \angleCBA = 25°.

So,

=> \angleCBD + \angleDCB = 180° (Angle sum Property.)

=> 90° +25° + \angleDCB = 180°

=> 115° + \angleDCB = 180°

=> \angleDCB = 180° - 115°

=> \angleDCB = 65°.

Now, in △ADC,

=> CD is the altitude to AB.

So,

=> \angleADC = 90°

=>\angle CAD =\angle CAB = 65°.

So,

=> \angleADC + \angleCAD +\angleDCA = 180° (Angle sum Property.)

=> 90° + 65° + \angleDCA = 180°

=> 155° +\angleDCA = 180°

=> \angleDCA = 180° - 155°

=> \angleDCA = 25°

Hence, we get,

  • \angleDCA = 25°
  • \angleDCB = 65°
  • \angleCDB = 90°
  • \angleACD = 25°
  • \angleADC = 90°.
7 0
3 years ago
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