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3 years ago

A farmer uses 3/7 of land and another 2/5 of land how much land is left

Mathematics

1 answer:

KonstantinChe [14]3 years ago

5 0

The farmer has used:
3/7 of land+ 2/5 of land= 29/35 of land.

1- 29/35= 6/35

6/35 of land is left~

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What is the solution to this 20x+30y=1000

PilotLPTM [1.2K]

x = −3/ 2 y + 50 is your answer.

If you would like me to explain, let me know I would be happy to help!

Hope this helped! Have a great day/afternoon!

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3 years ago

83,78, 99,56, 48, 74, 68,55,85

____ [38]

Answer:

48, 55, 56, 68, 74, 78, 83, 85, 99

Mean: 71.77777777777777777..........

Median: 74

Mode: mode is the number seen the most, but there is no repeating values

Range: 51

Step-by-step explanation:

Mean:

add all the values together, and divide by how many values you have, so 646/9= 71.777777777............

Median:

the number in the middle of the sequence

Range:

the difference between the largest and smallest numbers/values.

I hope this helps u pls give a brainliest and a thx ;)

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3 years ago

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Recently, Frank took a one-hundred question aptitude test where each correct answer scored 5 points, each incorrect answer score

Lyrx [107]

Im guessing he answered 30 correct, let me know if im correct

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3 years ago

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

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3 years ago

URGENT WILL GIVE BRAINLIEST ANSWER ASAP!

dmitriy555 [2]

3i = 105

To make i by itself, divide both sides by i.

i = 35

There you have it, 35 is the answer.

3 0

3 years ago

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