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pashok25 [27]
3 years ago
10

A plumber charges a one-time service fee of $20 in addition to his hourly

Mathematics
1 answer:
zalisa [80]3 years ago
8 0

Answer:

Three hundred fifty-five

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Caroline wants to buy a pair of shoes that has an original price of $40.00. She has a coupon for a 30% discount off of the origi
fredd [130]
Answer:
$26.32

step-by-step explanation:
it originally was $40.00
she has a coupon for 30%
30% of 40.00 is 12
40 • 0.30 = 12
40 - 12
$28
it is now $28 after the coupon
her state charges an 6% sales tax on the discounted price
6% of 28 is 1.68
28 • 0.06 = 1.68
28 - 1.68 = 26.32
6 0
3 years ago
The first pentagon is dilated to form the second pentagon. Drag and drop the answer to correctly complete the statement.
Hitman42 [59]
If the bigger pentagon has side length of 5 units and the smaller pentagon has side length of 1 unit, you can know that the dilation was 1/5 = 0.2

This means that the sides of the big pentagon were multiplied by 0.2 to obtain the smaller pentagon.
4 0
3 years ago
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What methods could you use to calculate the x-coordinate of the midpoint of a horizontal segment with the endpoints of (-20, 0)
Ronch [10]
The coordinates (abscissa and ordinates) of the midpoint of the line segment is the average of the abscissas and the ordinates, respectively. Therefore, the x - coordinate of the midpoint is equal to -20 + 20 or 0 by 2 which is zero. The answer is letter E. 
7 0
4 years ago
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Emily rented a bike from matt’s bikes. It cost $16 plus $5 per hour. If Emily paid $46 then she rented the bike for how many hou
Over [174]
16+5x=46
5x=30
x=6

She rented the bike for 6 hours
5 0
4 years ago
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A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the
amm1812

Answer:

s(t)=(ln|7|+ln|2|)\,ft

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\

Integrating both sides w.r.to t

v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\

Using initial value

v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}

We have to calculate the distance covered in time interval [0,5], so:

\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft

3 0
3 years ago
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