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Reika [66]
3 years ago
11

BRIANLIESTT!!!!GIVING

Mathematics
2 answers:
Bogdan [553]3 years ago
7 0

Answer:

A

Step-by-step explanation:

YES A BELEIVE MEH IT A YES YES A A A A A A IT IS A DEFINITELY A

balu736 [363]3 years ago
6 0

Answer:

A.

Step-by-step explanation:

x - 9 = 0

  + 9 + 9

---------------

x = 9

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At a 33-foot depth underwater, the pressure is 29.55 pounds per square inch. At a depth of 66 feet, the pressure reaches 44.4 po
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In Physics, the usual pattern or principle for pressure suggests that as the force increases and the area decreases, the pressure increases which is dependent as to the interval of both force and area's intensifies or diminishes. Though in this problem, if we base on the given data it seems that the pressure is currently increasing and is faced by a positive correlation at almost 50% in rate as evident that when the 33 ft increased to about 66ft the latter simultaneously rise with it.
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What is the price of a $13 ticket with a 8% sales tax?
HACTEHA [7]

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$14.04

Step-by-step explanation:

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5 0
3 years ago
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Find Area of A regular polygon. <br> A square with a side of 5?
Daniel [21]

Area of square is

A=a^2

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Hope this helps.

8 0
3 years ago
First to awnser will the question correct will get brainlyist
inessss [21]

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x 1.12

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5 0
4 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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