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3 years ago

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The builder receives a 20% contractor's discount, plus and additional 3% for paying in full within 30 days. Tax on building mate

rials is 5%. What is the total bill including discounts and taxes?

1 answer:

erastova [34]3 years ago

5 0

Answer:

$1340.49

Step-by-step explanation:

From the attached image

Subtotal Cost of Purchased Items

$= (20 \times 16.69)+(2 \times 20.78)+(2 \times 15.58)+(6 \times 21.38)\\+(118 \times 6.60)+(500 \times 0.22)+(10 \times 8.44)+(1 \times 150)\\\\=\$1658$

Since the builder receives a 20% contractor's discount, plus and additional 3% for paying in full within 30 days.

Discount = 23% of $1658 = 0.23 X 1658 =$381.34

Total (Less Discount) = 1658-381.34 = $1276.66

Tax on building materials is 5%.

Therefore:

Tax = 5% of $1276.66=0.05 X 1276.66

Tax=$63.83

Therefore, the total bill:

=$1276.66+63.83

Grand Total =$1340.49

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3 years ago

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Add [2] and [3]:<span>
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4 years ago

Read 2 more answers

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Marat540 [252]

Given:

The point, (4, -3)

The line,

$y=\frac{1}{2}x+5$

To find an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the given line:

The slope of the line is,

$m=\frac{1}{2}$

Since the given line is parallel to the new line, so the slope will be same for the both.

Using the point-slope formula,

$y-y_1=m(x-x_1)$

Substitute the point and slope we get,

$\begin{gathered} y-(-3)=\frac{1}{2}(x-4) \\ y+3=\frac{1}{2}x-2 \\ y=\frac{1}{2}x-2-3 \\ y=\frac{1}{2}x-5 \end{gathered}$

Hence, the equation in slope-intercept form for the line is,

$y=\frac{1}{2}x-5$

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2 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

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Answer:

<em>C.</em> $(5-\frac{1}{2})^6$

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;

$Sum = 2 + 4$

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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]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

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

$^nC_r = 15$

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$b^r= (-\frac{1}{2})^4$

[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 $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em> $(5-\frac{1}{2})^6$ <em />

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