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Mice21 [21]
2 years ago
5

A cashier earns $7 an hour. if x is the number of hours worked, which function represents the cashier’s earnings? y = 7x y = x s

uperscript 7 y = startfraction 7 over x endfraction y = 7
Mathematics
1 answer:
alisha [4.7K]2 years ago
3 0

Answer:

y=7x

Step-by-step explanation:

The job pays $7 per 1 hour so to find out how much money you get for x hours, you would multiply the hourly rate ($7) by the amount of hours (x)

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A triangle with an area of 23 cm² is dilated by a factor of 6.
Jlenok [28]
For this item, I will assume that we are required to give the area of the dilated triangle. By dilated we mean to say that the dimensions of the second triangle is 6 times that of the first. We square 6 and them multiply this to the area of the original triangle to get the area of the second. That is,
                     area of second triangle = (2/3 cm²)(6²) = 24 cm²
Thus, the area of the new triangle is equal to 24 cm².
8 0
4 years ago
NEED HELP!!!
Dahasolnce [82]

Answer:

No.

Step-by-step explanation:

The solution x = 0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation.

7 0
3 years ago
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LenaWriter [7]

Answer:

Hi there!

The correct answer is: Based on parent function, the graph vertically stretched by a factor of 2, shifting to the left by 1 unit and moving vertically down by 5.

3 0
3 years ago
51/60 as a fraction, decimal, and a percent?
raketka [301]
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7 0
3 years ago
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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