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Gnesinka [82]
3 years ago
7

Consider the equation y = 4x - 10. Find y if x = 3.

Mathematics
2 answers:
xenn [34]3 years ago
8 0

Answer:

The simplified system is the arbitrary solution of the original system of equations

Step-by-step explanation:

Y=2 and x=3

Correct me if I’m wrong.

Hope this helps

olganol [36]3 years ago
7 0

Answer:

y = 2

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

y = 4x - 10

x = 3

<u>Step 2: Evaluate</u>

  1. Substitute in <em>x</em>:                                                                                                y = 4(3) - 10
  2. Multiply:                                                                                                           y = 12 - 10
  3. Subtract:                                                                                                          y = 2
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U have 10 batteries sitting on your desk, three of which are dead. You choose 2 batteries are random for you calculator. What is
Nesterboy [21]
You have to calculate the expected value of pulling any number of good batteries.
There are 3 bad batteries (B) and seven good batteries (G)
If you pull two batteries the possible number of good batteries you can get are 0, 1 and 2.
GB, BG, GG, and BB
two chances for getting 1, one chance for getting two, and one chance for getting zero.
In order to calculate the expected value you have to first calculate the values of all the possibilities.
(GB = 7/10 x 3/10) (BG = 3/10 x 7/10) (GG = 7/10 x 7/10) (BB = 3/10 x 3/10)
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3 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
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tiny-mole [99]

Answer:

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2500\times \frac{1}{5}=500

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