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2 years ago

How many solutions can be found for the equation 5x + 3(x − 1) = 10x − 2x − 3? (4 points) None

Mathematics

1 answer:

zzz [600]2 years ago

8 0

Infinitely many is the correct answer

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Select the statement that describes the following expression 6 x (2 + 7) − 9. (4 points)

stiv31 [10]

Answer:

Add all of them together

Step-by-step explanation:

You have to add the numbers and that gives you the sum

8 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

What is f(2) if f(x) = 2x-5 and f(4) if f(x) = -3x + 15 step by step please!

mash [69]

Answer:

f(2) = - 1 and f(4) = 3

Step-by-step explanation:

To evaluate f(2), substitute x = 2 into f(x), that is

f(2) = 2(2) - 5 = 4 - 5 = - 1

To evaluate f(4), substitute x = 4 into f(x)

f(4) = - 3(4) + 15 = - 12 + 15 = 3

8 0

3 years ago

Simplify 13x + 7y - 2x + 6a

VLD [36.1K]

Step-by-step explanation:

13x+7y-2x+6a

combine like terms

=13x-2x+7y+6a

=11x+7y+6a

simplified

8 0

3 years ago

Read 2 more answers

Which of the following statements is false?

AleksAgata [21]

Statement C is false.

7 0

4 years ago