Ask question

Ask question

All categories

3 years ago

8

Place 16 bingo chips in 4 rows of 4 each. Remove 6 bingo chips, leaving an even number of bingo chips in each row and column. (E

ven means an even number such as 2 or 4)

1 answer:

topjm [15]3 years ago

4 0

Answer:this would leave 10 bingo chips left.

Step-by-step explanation: start with 4 rows of 4 chips, take away 6 chips leaves 10

You might be interested in

Solve for x<br> 6/x^2+2x-15 +7/x+5 =2/x-3

timama [110]

Answer:

x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 17/(3 (10700 - 45 sqrt(56235))^(1/3)) - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3))

Step-by-step explanation:

Solve for x:

6/x^2 + (2 x - 8)/(x + 5) = 2/x - 3

Bring 6/x^2 + (2 x - 8)/(x + 5) together using the common denominator x^2 (x + 5). Bring 2/x - 3 together using the common denominator x:

(2 (x^3 - 4 x^2 + 3 x + 15))/(x^2 (x + 5)) = (2 - 3 x)/x

Cross multiply:

2 x (x^3 - 4 x^2 + 3 x + 15) = x^2 (2 - 3 x) (x + 5)

Expand out terms of the left hand side:

2 x^4 - 8 x^3 + 6 x^2 + 30 x = x^2 (2 - 3 x) (x + 5)

Expand out terms of the right hand side:

2 x^4 - 8 x^3 + 6 x^2 + 30 x = -3 x^4 - 13 x^3 + 10 x^2

Subtract -3 x^4 - 13 x^3 + 10 x^2 from both sides:

5 x^4 + 5 x^3 - 4 x^2 + 30 x = 0

Factor x from the left hand side:

x (5 x^3 + 5 x^2 - 4 x + 30) = 0

Split into two equations:

x = 0 or 5 x^3 + 5 x^2 - 4 x + 30 = 0

Eliminate the quadratic term by substituting y = x + 1/3:

x = 0 or 30 - 4 (y - 1/3) + 5 (y - 1/3)^2 + 5 (y - 1/3)^3 = 0

Expand out terms of the left hand side:

x = 0 or 5 y^3 - (17 y)/3 + 856/27 = 0

Divide both sides by 5:

x = 0 or y^3 - (17 y)/15 + 856/135 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

x = 0 or 856/135 - 17/15 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

x = 0 or z^6 + z^4 (3 λ - 17/15) + (856 z^3)/135 + z^2 (3 λ^2 - (17 λ)/15) + λ^3 = 0

Substitute λ = 17/45 and then u = z^3, yielding a quadratic equation in the variable u:

x = 0 or u^2 + (856 u)/135 + 4913/91125 = 0

Find the positive solution to the quadratic equation:

x = 0 or u = 1/675 (9 sqrt(56235) - 2140)

Substitute back for u = z^3:

x = 0 or z^3 = 1/675 (9 sqrt(56235) - 2140)

Taking cube roots gives (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) times the third roots of unity:

x = 0 or z = (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) or z = -((-1)^(1/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) or z = ((-1)^(2/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3))

Substitute each value of z into y = z + 17/(45 z):

x = 0 or y = (9 sqrt(56235) - 2140)^(1/3)/(3 5^(2/3)) - (17 (-1)^(2/3))/(3 (5 (2140 - 9 sqrt(56235)))^(1/3)) or y = 17/3 ((-1)/(5 (2140 - 9 sqrt(56235))))^(1/3) - ((-1)^(1/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) or y = ((-1)^(2/3) (9 sqrt(56235) - 2140)^(1/3))/(3 5^(2/3)) - 17/(3 (5 (2140 - 9 sqrt(56235)))^(1/3))

Bring each solution to a common denominator and simplify:

x = 0 or y = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) or y = 1/15 (17 5^(2/3) ((-1)/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) or y = -(2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3)) - 17/(3 (5 (2140 - 9 sqrt(56235)))^(1/3))

Substitute back for x = y - 1/3:

x = 0 or x = 1/15 (2140 - 9 sqrt(56235))^(-1/3) ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 1/3 5^(-2/3) (2140 - 9 sqrt(56235))^(1/3) - 17/3 (5 (2140 - 9 sqrt(56235)))^(-1/3)

5 (2140 - 9 sqrt(56235)) = 10700 - 45 sqrt(56235):

x = 0 or x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3)) - 17/(3 (10700 - 45 sqrt(56235))^(1/3))

6/x^2 + (2 x - 8)/(x + 5) ⇒ 6/0^2 + (2 0 - 8)/(5 + 0) = ∞^~

2/x - 3 ⇒ 2/0 - 3 = ∞^~:

So this solution is incorrect

6/x^2 + (2 x - 8)/(x + 5) ≈ -3.83766

2/x - 3 ≈ -3.83766:

So this solution is correct

6/x^2 + (2 x - 8)/(x + 5) ≈ -2.44783 + 1.13439 i

2/x - 3 ≈ -2.44783 + 1.13439 i:

So this solution is correct

6/x^2 + (2 x - 8)/(x + 5) ≈ -2.44783 - 1.13439 i

2/x - 3 ≈ -2.44783 - 1.13439 i:

So this solution is correct

The solutions are:

Answer: x = ((-5)^(1/3) (2140 - 9 sqrt(56235))^(2/3) - 17 (-5)^(2/3))/(15 (2140 - 9 sqrt(56235))^(1/3)) - 1/3 or x = 1/15 (17 5^(2/3) (-1/(2140 - 9 sqrt(56235)))^(1/3) - (-5)^(1/3) (9 sqrt(56235) - 2140)^(1/3)) - 1/3 or x = -1/3 - 17/(3 (10700 - 45 sqrt(56235))^(1/3)) - (2140 - 9 sqrt(56235))^(1/3)/(3 5^(2/3))

4 0

2 years ago

Find the value of x..........

Leviafan [203]

Answer:

x = 53.5

Step-by-step explanation:

x is one half the difference of the intercepted arcs, that is

x = 0.5(137 - 30) = 0.5 × 107 = 53.5

8 0

2 years ago

Are you in collage because you wont get the question unless you are lol

Troyanec [42]

Are you in college because you misspelled college

4 0

2 years ago

A sample of nitrogen in a glass bulb has a mass of 245 mg. What is this mass in kilograms? (Note: 1,000 mg = 1 g, 1,000 g = 1 kg

notka56 [123]

You can use the specified conversion scale to convert the weight given to kilograms.

The mass of given bulb in kilogram is 0.000245 kilogram.

<h3>How to convert from one unit to another unit?</h3>

If a quantity is given in one unit and its unit of measurement has to be changed to other unit, then you will have to get to know how is first unit related to other. Based on that, you can convert the unit of measurement to desired unit of measurement.

<h3>How to convert 245 mg to kilogram?</h3>

Since 1000 mg = 1 gram, thus,

Dividing by 1000 on both sides, we get:

$\dfrac{1000}{1000} \: \rm mg = \dfrac{1}{1000} \: \rm g\\
\\
1 \: mg = 0.001 \: g\\
\\
\text{Multiplying 245 on both sides}\\\\
245 \times 1 \: mg = 245 \times 0.001 \: g = 0.245 \: g\\
245 \: mg = 0.245 \: g$

Now since 1000g = 1 kg, thus,

Dividing by 1000 on both the sides,

$\dfrac{1000}{1000} \: \rm g = \dfrac{1}{1000} \: \rm kg\\\\1 \: g = 0.001 \: kg\\\\\text{Multiplying 0.245 on both sides}\\\\0.245 \times 1 \: g = 0.245 \times 0.001 \: kg = 0.000245 \: kg\\0.245 \: g = 0.000245 \: kg$ $\dfrac{1000}{1000} \: \rm g = \dfrac{1}{1000} \: \rm kg\\\\1 \: g = 0.001 \: kg\\\\\text{Multiplying 0.245 on both sides}\\\\0.245 \times 1 \: g = 0.245 \times 0.001 \: kg = 0.000245 \: kg\\0.245 \: g = 0.000245 \: kg$

Thus, 245 mg = 0.245g = 0.000245 kg

The mass of given bulb in kilogram is 0.000245 kilogram.

Learn more about unit conversion here:

brainly.com/question/16418622

8 0

2 years ago

If Tom swam 1000 me in 2/3 of an hour how fast did he swim

vekshin1

Step-by-step explanation:

one hour=60min

2/3 of an hour

=2/3x60=40min

1000 metres in 40 minutes

4 0

3 years ago