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3 years ago

Select the correct answer.

Mathematics

1 answer:

jarptica [38.1K]3 years ago

7 0

A is the correct answer

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Irene can snap 48 times in 8 minutes at a constant rate. How many

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6 snaps a minute if you multiply 9 x 6 is 54

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Divide 2.875x10^6by2.3x10^3

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1875L = 1.875x10^6 mL

multiply by density to get 2.14x10^6

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Which equation is represented by the graph below?

LuckyWell [14K]

Answer:

Step-by-step explanation:

Hello

A (5;2)

B (-1;-7)

y = ax + b

2 = 5a + b

-7 = -a + b

2 - (-7) = 5a - (-a) + b - b

2 + 7 = 5a + a

9 = 6a

a = 9/6

a = 3/2

2 = 3/2 * 5 + b

b = 2 - 15/2

b = 4/2 - 15/2

b = 11/2

y = 3/2 x + 11/2

y+1= 3/2(x-4)

y + 1 = 3/2 x - 6

y = 3/2 x - 6 - 1

y = 3/2 x - 7 => no

y-4= 3/2(x+1)

y - 4 = 3/2 x + 3/2

y = 3/2 x + 3/2 + 4

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y = 3/2 x + 11/2 => yes

y+4= 3/2(x-1)

y + 4 = 3/2 x - 3/2

y = 3/2 x - 3/2 - 4

y = 3/2 x - 3/2 - 8/2

y = 3/2 x - 11/2 => no

y-1= 3/2 (x+4)

y - 1 = 3/2 x + 6

y = 3/2 x + 6 + 1

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Solve the inequality and graph its solution... 3+v≤-9

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Answer:

v≤-12, and the graph is left 12

Step-by-step explanation:

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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