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3 years ago

PLEASE HELP ASAP SHOW WORK PLEASE!!!!

Mathematics

1 answer:

gregori [183]3 years ago

3 0

Answer:

The limit is -1

Step-by-step explanation:

As shown by the equation, it shows x -> -1

which represents x approaching the limit of -1

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Almost done with my hw but stuck on this one please help

Mashutka [201]

Answer:

Because that angle is 30°.

it's nearest value is 27.

6 0

3 years ago

Rose works eight hours a day for five days a week. how many hours will she work in six weeks ?

Kryger [21]

Start by multiplying 8 and 5 to get the hours worked each week.
Next multiply that by 6 to get how many hours she'll work in 6 weeks.

7 0

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

3 0

3 years ago

A certain number of sixes and nines is added to give a sum of 126; if the number of sixes and nines is interchanged, the new sum

galina1969 [7]

Answer:

Original number of sixes = 6

Original number of nines = 10

Step-by-step explanation:

We are told in the question that:

A certain number of sixes and nines is added to give a sum of 126

Let us represent originally

the number of sixes = a

the number of nines = b

Hence:

6 × a + 9 × b = 126

6a + 9b = 126.....Equation 1

If the number of sixes and nines is interchanged, the new sum is 114.

For this second part, because it is interchanged,

Let us represent

the number of sixes = b

the number of nines = a

6 × b + 9 × a = 114

6b + 9a = 114.......Equation 2

9a + 6b = 114 .......Equation 2

6a + 9b = 126.....Equation 1

9a + 6b = 114 .......Equation 2

We solve using Elimination method

Multiply Equation 1 by the coefficient of a in Equation 2

Multiply Equation 2 by the coefficient of a in Equation 1

6a + 9b = 126.....Equation 1 × 9

9a + 6b = 114 .......Equation 2 × 6

54a + 81b = 1134 ........ Equation 3

54a + 36b = 684.........Equation 4

Subtract Equation 4 from Equation 3

= 45b = 450

divide both sides by b

45b/45 = 450/45

b = 10

Therefore, since the original the number of nines = b,

Original number of nines = 10

Also, to find the original number of sixes = a

We substitute 10 for b in Equation 1

6a + 9b = 126.....Equation 1

6a + 9 × 10 = 126

6a + 90 = 126

6a = 126 - 90

6a = 36

a = 36/6

a = 6

Therefore, the original number of sixes is 6

7 0

3 years ago

On any given day, Juan will break even if he earns as much money as he spends. On day 5, he records the expression 16 + (–16). D

Degger [83]

Yes she do break even because let me show you a expression now take a number line and then you start at -16 right and then you go down the number line and then this is how it go -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 and i just answer your Q there you go -16+16=0

7 0

3 years ago

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