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3 years ago

What is 4\7 plus 1 1\3

2 answers:

maks197457 [2]3 years ago

7 0

The answer is 40/21 or 1 19/21 as a mixed number .
4/7+1 1/3=
1 1/3=4/3
4/7+4/3=
4/7 times 3/3= 12/21
4/3 times 7/7= 28/21
The common denominator is 21 , I multiplied the fractions by 3 and by 7 because 3 times 7= 21 and 7 times 3=21.
Answer: 40/21

Alika [10]3 years ago

3 0

You would get 89/21 and then for a mixed number 4& 5/21

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Verdich [7]

Answer:

I assume Josh and Peyton are his friends and both gave him advice on what to do with half of the money from the big lottery win.

Let's say Josh said "save it or invest it for your retirement" and Peyton said "use it to keep playing the lottery.

We will now look at the sense in each piece of advice!

Step-by-step explanation:

JOSH

By investing the $250,000 (half of the money won), Patrick will be sure that the money is available for him anytime and would even have gotten interest, by the time he's ready to use it.

PEYTON

By playing the lottery continuously, Patrick could get lucky once in a while and win big again. How big though?

Analyzing with the figures given,

$20 gets Patrick a lottery ticket.

$250,000 will get him 12,500 lottery tickets!

Whether he's buying the tickets at once or he'll play the lottery once in a while, I'll say he has good chances of winning big again.

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5 0

3 years ago

I'VE ASKED THIS 4 TIMES ALREADY. DOUBLE POINTS PLEASE HELP!!!

astraxan [27]

Answer:

y=-3x-14

Step-by-step explanation:

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2 years ago

What expression simplifies to 2i ? help asap lots of points

PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

First, consider the following expression.

(6x + 8) + (4x + 2)

To simplify this expression, you combine the like terms, 6x and 4x. These are like terms because they have the same variable with the same exponents. Similarly, 8 and 2 are like terms because they are both constants, with no variables.

(6x + 8) + (4x + 2) = 10x + 10

5 0

3 years ago

Read 2 more answers

describe how to transform (^5 square root x^7)^3 into an expression with a rational exponent. make sure you respond with complet

diamong [38]

Answer:

$x^\frac{21}{5}$

Step-by-step explanation:

We are given an expression and we have to transform it into an expression with exponent.

5th root of x can be written as $x^\frac{1}{5}$

$(\sqrt[5]{x^7})^3\\\\((x^7)^\frac{1}{5} )^3$

The powers outside will be multiplied as: $(x^a)^b = x^{ab}$

$(x^\frac{7}{5} )^3\\x^\frac{7*3}{5}\\x^\frac{21}{5}$

where the exponent is $\frac{21}{5}$ and it is a rational number by definition of rational numbers

3 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