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3 years ago

Age Problem. Please help. I suck at these.

1 answer:

7 0

Let Cynthia's age be C, then Abigail's age A

since Abigail is 8 years older, her age A = C+8.

20 years ago Cynthia's she was C-20 and Abigail was A-20=3(C-20)

A-20=3C-60
A=3C-40
but we just said A=C+8
so 3C-40=C+8
2C=48
C=24
A=24+8=32

Cynthia's age is 24 and Abigail's age is 32

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Use the decimals 2.47, 9.57 and 7.1 to write two different addition facts and two different subtraction facts

netineya [11]

2.47 + 7.1
7.1 + 2.47

9.57-7.1
9.57-2.47

6 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

-12-x>-12+4x put the steps too pls thanks

andrey2020 [161]

Answer:

x < 0

Step-by-step explanation:

Firstly put the values with variable x to one side,

; -12 - x > -12 + 4x

; -12 + 12 > 4x + x

; 0 > 5x

then divide both sides by 5,

; (0 ÷ 5) > (5x ÷ 5)

; 0 > x

; Hence the solution to the inequality above,

; x < 0...always remember to reverse the sign when you change the side to which you put the variable of x,

Therefore...( x < 0 )

3 0

3 years ago

X-y=-3 Anyone know how to do this ?

horsena [70]

X - y = -3

Move constant to the left by adding its opposite to both sides

x - y + 3 = -3 + 3

The sum of two opposites equals

your answer would be: x - y + 3 = 0

7 0

2 years ago

The point on the number line is located at which of the following mixed numbers?

I am Lyosha [343]

Answer:

is 6 3/4

Step-by-step explanation:

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3 years ago

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