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3 years ago

Is 5 1/2 bigger then 5 3/8

Mathematics

2 answers:

Len [333]3 years ago

8 0

This is true because 5 1/2 translates to 5 4/8 which is greater than 5 3/8.

Liono4ka [1.6K]3 years ago

6 0

1/2 and 3/8

equivalent

1/2 = 4/8

so 4/8 is greater than 3/8

Answer

5 1/2 is bigger than 5 3/8

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Most of the numbers we know, and work with, are Real Numbers. The Real Number System includes counting numbers, fractions, terminating decimals, positive numbers, negative numbers, zero, repeating decimals, never ending and non-repeating decimals, numbers that are expressed as radicals, and even pi (π).

The natural numbers are the set of counting numbers

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The multiplication of two natural numbers creates another natural number.

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BUT ...

The subtraction of two natural numbers does NOT necessarily create another natural number

The division of two natural numbers does NOT necessarily create another natural number

The whole numbers are the set of counting numbers (natural numbers) along with zero

• There are infinitely many numbers in this set of numbers.

• The set of whole numbers is "closed" under addition and multiplication.

Integers:

• The integers are the set of all of the natural numbers,

plus their additive inverses and zero

• The integers are "closed" under addition, multiplication and subtraction,

but NOT under division

Rational Numbers:

• The rational numbers are the set of numbers which can be expressed as a ratio

(a fraction) between two integers.

• Integers are rational numbers since 5 can be written as the fraction 5/1.

• Decimals which terminate are rational numbers.

• Decimals which have a repeating pattern are rational numbers. 1/3 = 0.3333333...

• The rational numbers are "closed" under addition, subtraction, and multiplication. Under division, we run into the problem of division by 0, which makes the statement that "the rationals are closed under division" false. Some texts state that "the rationals are closed under division as long as the division is not by zero" which is a true statement.

Irrational Numbers:

The irrational numbers are the set of number which can NOT be written as a ratio (fraction).

• Decimals which never end nor repeat are irrational numbers.

• Irrational numbers are "not closed" under addition, subtraction, multiplication or division.

• Examples of irrational numbers: rad2, π

8 0

3 years ago

Read 2 more answers

Determine the number of real solutions of -2x^2+5x-3=0

Tema [17]

Answer:

Two distinct real solutions.

Step-by-step explanation:

Given the equation in the form $ax^2+bx+c=0$ , you need to find the Discriminant with this formula:

$D=b^2-4ac$

For the equation $-2x^2+5x-3=0$ you can identify that:

$a=-2\\b=5\\c=-3$

Then, substituting these values into the formula, you get that the Discriminant is:

$D=5^2-4(-2)(-3)$

$D=1$

Since $D>0$ , then $-2x^2+5x-3=0$ has two distinct real solutions.

8 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago