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Klio2033 [76]
4 years ago
14

Please help me !!!!!!!

Mathematics
1 answer:
matrenka [14]4 years ago
4 0
I THINK ITS D HAVE A GOOD DAY
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When testing for current in a cable with eight ​color-coded wires, the author used a meter to test five wires at a time. How man
balu736 [363]

Answer:

56 different tests

Step-by-step explanation:

Given:

Number of wires available (n) = 8

Number of wires taken at a time for testing (r) = 5

In order to find the number of different tests required for every possible pairing of five wires, we need to find the combination rather than their permutation as order of wires doesn't disturb the testing.

So, finding the combination of 5 pairs of wires from a total of 8 wires is given as:

^nC_r=\frac{n!}{r!(n-r)!}

Plug in the given values and solve. This gives,

^8C_5=\frac{8!}{5!(8-5)!}\\\\^8C_5=\frac{8\times 7\times 6\times 5!}{5!\times 3\times2\times1}\\\\^8C_5=56

Therefore, 56 different tests are required for every possible pairing of five ​wires.

6 0
3 years ago
This Question: 1 pt
shtirl [24]
50 miles apart
if 1 = 10 then 5 = 50 using the tens rule
3 0
3 years ago
36x+32=32x-36<br> whats the value of x
bulgar [2K]

Answer:

x = -17

Step-by-step explanation:

<u>Step 1:  Subtract 32x from both sides</u>

36x + 32 - 32x = 32x - 36 - 32x

4x + 32 = -36

<u>Step 2:  Subtract 32 from both sides</u>

<u />4x + 32 - 32 = -36 - 32

4x = -68

<u>Step 3:  Divide both sides by 4</u>

<u />4x / 4 = -68 / 4

x = -17

Answer:  x = -17

8 0
4 years ago
Please help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
IRISSAK [1]

Answer:

4 packages

Step-by-step explanation:

450 can be divided by 100 4 whole times, the extra cards cannot be added into a package

7 0
3 years ago
Read 2 more answers
Determine the number of real solutions each quadratic equation has. y = 12x2 - 9x 4 real solution(s) 10x y = -x2 2 real solution
nordsb [41]

All the given equations have 2 real solutions except for equation 4 which has only one real root.

<h3>What is a quadratic equation?</h3>

A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is  ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.

The given quadratic equation

y = 12x^2 - 9x+ 4

where, a = 12, b = -9 and c = 4

x = -b + \sqrt{b^2 - 4ac} /2a

x = -(-9)+ \sqrt{(-9)^2 - 4(12)(4)} /2(12)

x = 1.063 or -0.313

Second Equation

10x +y = -x^2 +2 \\y = x^{2} +10x - 2

where, a = 1, b = 10, c = -2

substitute the values into the equation

x = -b + \sqrt{b^2 - 4ac} /2a

-10 + \sqrt{10^2 - 4(1)(-2)} /2(1)

x = 0.195 or -10.195

Third Equation

4y - 7 = 5x^2 - x +2 +3y

y = 5x^{2} -x -5

where, a = 5, b = -1, c = - 5

x = -b + \sqrt{b^2 - 4ac} /2a

-(-1) + \sqrt{(-1)^2 - 4(5)(-5)} /2(5)

x = 0.905 or -0.55

Fourth Equation

y = (-x +4)^2 \\y = x^{2} -8x +16

where, a = 1, b = -8, c = 16

x = -b + \sqrt{b^2 - 4ac} /2a

x = -(-8) + \sqrt{(-8)^2 - 4(1)(16)} /2(1)

x = 4

Learn more about quadratic equations;

brainly.com/question/13197897

6 0
2 years ago
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