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4 years ago

A number x is both less than 1.5 and greater than or equal to 0

1 answer:

5 0

$x$

$x\ge~0$

Another way to write this is:

$0\le~x$

So 'x' can be any number between 0 and 1.5, not including 1.5, but including 0.

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3 years ago

Solve the system of equations.<br> y=x+2<br> y=4x+23<br> Write your answer as<br> X=<br> y=

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Answer:

Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Please I’ve attached the question can you help me

Andreyy89

Answer:

1. <em>n</em>[(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. <em>n</em>[(A - B) U C] = <em>n</em>[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 <em>n</em>[(A U B) - C]

<em>n</em>(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}

<em>n</em>[(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 <em>n</em>(A U B) and C, we therefore remove them it and have:

<em>n</em>[(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 <em>n</em>[(A - B) U C]

To get <em>n</em>(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:

<em>n</em>[(A - B) U C] = <em>n</em>[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

3 years ago

How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the

miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

$\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}$

___

$\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}$

___

$\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}$

Neither solution makes any denominator be zero, so both are good solutions.

8 0

3 years ago