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3 years ago

7

Sean Orr’s gross power bill for July was $189.76. By paying early he earned a discount of $3.79. What was the net amount of his

bill?

1 answer:

USPshnik [31]3 years ago

8 0

Sean’s power due to the discount will cost $186.10

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What are the zeros of the function f(x) = x2 – 10x + 21?<br> help bro

Nady [450]

Answer:

3, 7

Step-by-step explanation:

"Zeros" means solutions to the equation f(x) = 0. In other words, what are the values of x that make the function's value equal to zero?

This one can be solved by factoring.

$x^2-10x+21=0\\(x-7)(x-3)=0\\x-7=0 \text{ or } x-3=0\\x=7 \text{ or } x=3$

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3 years ago

Every evening Jenna empties her pockets and puts her change in a jar. At the end of the week she counts her money. One week she

lisov135 [29]

d = dimes

q = quarters

d+q = 38 coins

q=38-d

0.25q + 0.10d=6.95

0.25(38-d)+0.10d=6.95

9.5-0.25d+0.10d=6.95

-015d=-2.55

d=-2.55/-0.15 = 17

q=38-17 =21

21*0.25 =5.25

17*0.10 = 1.70

5.25+1.70 = 6.95

she had 21 quarters and 17 dimes

7 0

3 years ago

Read 2 more answers

◕ A dealer sold a photocopy machine at Rs 4200 with 13% VAT to a retailer. The retailer added transportation cost of Rs 250 , pr

satela [25.4K]

<u>Given </u><u>:</u><u>-</u>

A dealer sold a photocopy machine at Rs 4200 with 13% VAT to a retailer.
The retailer added transportation cost of Rs 250 , profit Rs 300 and local tax Rs 150 and sold to consumer .
Customer has to pay 13% VAT .

<u>To </u><u>Find </u><u>:</u><u>-</u>

Amount to be paid by the customer .

<u>Sol</u><u>u</u><u>tion </u><u>:</u><u>-</u>

Here , according to the question ,

$\sf\dashrightarrow Cost\ of \ machine \ = \ Rs 4200$

$\sf\dashrightarrow VAT = 13\%$

Therefore cost after adding VAT ,

$\sf\dashrightarrow Cost = Cost_{initial}+13\% of Cost_{initial}\\$

$\sf\dashrightarrow Cost = Rs4200 + Rs 4200 \times \dfrac{13}{100}\\$

$\sf\dashrightarrow Cost = Rs4200 + Rs546\\$

$\sf\dashrightarrow \bf{ Cost = Rs 4746}$

Again the values added by the retailer before selling to customer ,

Transport = Rs 250
Profit = Rs 300
Tax = Rs 150

Therefore total cost after adding these ,

$\sf\dashrightarrow Cost = Rs( 4746+ 250+ 300 + 150 )\\$

$\sf\dashrightarrow\bf { Cost = Rs 5446}$

Again Selling price after addition of 13% VAT ,

$\sf\dashrightarrow SP = Rs 5446 + 13\% \ of \ Rs 5446\\$

$\sf\dashrightarrow SP = Rs 5446 + \dfrac{13}{100}\times 5446 \\$

$\sf\dashrightarrow SP = Rs (5446 + 707.98) = Rs ( 5446 + 708)\\$

$\sf\dashrightarrow \underline{\underline{\bf SP_{to\ the \ customer} = Rs 6154}}$

<u>Hence </u><u>the </u><u>amount </u><u>to </u><u>be </u><u>paid </u><u>by </u><u>the </u><u>customer </u><u>is </u><u>Rs </u><u>6</u><u>1</u><u>5</u><u>4</u><u> </u><u>.</u>

3 0

2 years ago

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Suppose that f(t) is continuous and twice-differentiable for t≥0. Further suppose f″(t)≥9 for all t≥0 and f(0)=f′(0)=0. Using th

Paha777 [63]

Give me a heart and I'll tell you the answer

5 0

3 years ago

The figure shown below is composed of a semicircle and a non-overlapping equilateral triangle and contains a hole that is also c

Sloan [31]

Check the picture below.

since we know the radius of the larger semicircle is 8, thus its diameter is 16, which is the length of one side of the equilateral triangle. We also know the smaller semicircle has a radius of 1/3, and thus a diameter of 2/3, namely the lenght of one side of the small equilateral triangle.

now, if we just can get the area of the larger figure and the area of the smaller one and subtract the smaller from the larger, we'll be in effect making a hole/gap in the larger and what's leftover is the shaded figure.

$\bf \stackrel{\textit{area of a semi-circle}}{A=\cfrac{1}{2}\pi r^2\qquad r=radius}~\hspace{10em}\stackrel{\textit{area of an equilateral triangle}}{A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\stackrel{side's}{length}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\left[ \stackrel{\textit{larger figure}}{\cfrac{1}{2}\pi 8^2~~+~~\cfrac{16^2\sqrt{3}}{4}} \right]\qquad -\qquad \left[ \cfrac{1}{2}\pi \left( \cfrac{1}{3} \right)^2 +\cfrac{\left( \frac{2}{3} \right)^2\sqrt{3}}{4}\right]}$

$\bf \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\frac{4}{9}\sqrt{3}}{4} \right] \\\\\\ \left[ 32\pi +64\sqrt{3} \right]\qquad -\qquad \left[ \cfrac{\pi }{18}+\cfrac{\sqrt{3}}{9} \right]~~\approx~~ 211.38 - 0.37~~\approx~~ 211.01$

3 0

3 years ago