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2 years ago

Write the expanded from of h(3k-12.4)

Mathematics

1 answer:

rewona [7]2 years ago

7 0

Answer:

3kh - 12.4h

Step-by-step explanation:

h(3k-12.4)

= h(3k) - h(12.4)

= 3kh - 12.4h

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Reciprocal of. 0×7/11

Tpy6a [65]

Answer:

it doesn't exist

Step-by-step explanation:

the expression 0×7/11 is equivalent to 0. 1/0 isn't possible, so its reciprocal doesn't exist.

8 0

2 years ago

Y = -x+5y=−x+5y, equals, minus, x, plus, 5

Serga [27]

The answer is D
Hope this hel

6 0

2 years ago

Which points are located in the third quadrant?

8090 [49]

Answer: for example (-2,-2), (-3,-6), and (-8,-10)

Step-by-step explanation: every point in the third quadrant is both negative.

8 0

1 year ago

Solve the equation in the complex number system. x^2-12x+40=0

mylen [45]

solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Step-by-step explanation:

Solve the equation:

$x^2-12x+40=0$

We can solve using quadratic Formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b = -12, c = 40

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(40)}}{2(1)}\\x=\frac{12\pm\sqrt{144-160}}{2}\\x=\frac{12\pm\sqrt{-16}}{2}\\We\,\,know\,\,\sqrt{-1}=i\\ x=\frac{12\pm4i}{2}\\ x=\frac{12+4i}{2}\,\,and\,\, x=\frac{12-4i}{2}\\ x=\frac{4(3+i)}{2}\,\,and\,\, x=\frac{4(3-i)}{2}\\x=2(3+i)\,\,and\,\,x=2(3-i)\\x=6+2i\,\,and\,\,x=6-2i$

So, solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Keywords: Solving quadratic equation

Learn more about Solving quadratic equation at:

#learnwithBrainly

5 0

2 years ago

Please I’ve attached the question can you help me

Andreyy89

Answer:

1. n[(A U B) - C] = {23, 24, 27, 29, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 61, 63}

2. n[(A - B) U C] = n[A U C] = {6, 10, 12, 15, 20, 23, 29, 30, 31, 37, 41, 43, 47, 53, 59, 60, 61}

3. D. I, II and III.

Step-by-step explanation:

U = {21, 22, 23, ..., 64}

A prime number is a number that can be divided only by 1 and itself.

A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

B = {24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63}

C = {6, 10, 12, 15, 20, 30, 60}

1. Find n[(A U B) - C]

n(A U B) = {23, 24, 27, 29, 30, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 60, 61, 63}

n[(A U B) - C] = {23, 24, 27, 29, 30, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 60, 61, 63} - {6, 10, 12, 15, 20, 30, 60}

Since only 30 and 60 are common to n(A U B) and C, we therefore remove them it and have:

n[(A U B) - C] = {23, 24, 27, 29, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 61, 63}

2. Find n[(A - B) U C]

To get n(A - B) we remove all the elements of B in A. Since there are no common elements between A and B, we therefore have:

A - B = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

C = {6, 10, 12, 15, 20, 30, 60}

Therefore, we have:

n[(A - B) U C] = n[A U C] = {6, 10, 12, 15, 20, 23, 29, 30, 31, 37, 41, 43, 47, 53, 59, 60, 61}

3. Which of the following is/are true?

I. A ∩ B = A ∩ C

A ∩ B = ∅

A ∩ C = ∅

Therefore, A ∩ B = A ∩ C is true.

II. A - B = A - C

A - B = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

A - C = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

Therefore, A - B = A - C is true.

III. A ∩ (B ∪ C) = ∅

A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

B U C = {6, 10, 12, 15, 20, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63}

Therefore, A ∩ (B ∪ C) = ∅ is true.

Therefore, the correct option is D i.e. I, II and III are true.

4 0

2 years ago