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2 years ago

3 - 12[(x + y) - 2)

Mathematics

1 answer:

agasfer [191]2 years ago

3 0

Answer:

Well plz dont delete my question I need points to get mine out

Step-by-step explanation:

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What is the standard form of this equation of a circle?

Olenka [21]

x² + y² + 14x − 4y − 28 = 0

x² +14x +y² - 4y =28
x²+2*7x +7² -7² + y² - 2*2y +2² - 2² = 28
(x+7)² + (y-2)² -7²-2² =28
(x+7)² + (y-2)²=28+49+4

(x+7)² + (y-2)² =81 is the answer.

5 0

3 years ago

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Ok Please Graph x > 2

Anon25 [30]

Answer:

Step-by-step explanation:

4 0

3 years ago

The median of a set of consecutive odd integers is 138. If the greatest integer in the set is 145, what is the least integer in

garri49 [273]

Answer:

131

Step-by-step explanation:

Given that:

Median of a consecutive odd integers = 138

Greatest integer In dataset = 145

Least integer in dataset =

Since the integers are consecutive ;

Number of integers to greatest integer value : Greatest integer - median value = 145 - 138 = 7

Hence, 7 integers before the median value should give us the least integer value ;

138 - 7 = 131

CHECK:

131,132,133,134,135,136,137,138,139,140,141,142,143,144,145

Median = (n+1)/2 th term

Median = (15+1)/2 = 8th term = 138

Highest = 145

Least = 131

7 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

2x-14=-3x+6 help with this equation???

Sphinxa [80]

2x-14=-3x+6
You need to combine like terms and get all the x's on one side and the numbers on the other
2x+3x-14=6
5x=6+14
5x=20
now divide to get x alone
x=4

Hope that helps

5 0

4 years ago

Read 2 more answers