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Anika [276]
3 years ago
6

WILL GIVE BRAINLIEST

Mathematics
2 answers:
jasenka [17]3 years ago
8 0

Answer:

a.1) well we can transform f(x) to g(x) by doing rise over run which is a simple way to find a linear function and transfer linear functions.

a.2) also when we pass the points givein if both f(x) and g(x) are part of the line that is draw to connect the points thats another way.

b) k is a variable just like the others therefore it can be divided by the 2 variables whivh you getx for both.

c) f(x) / g(x)----------------f(x) over g(x)

Step-by-step explanation:

i really hope this helps and brainliest if can welcome:)

Zina [86]3 years ago
5 0
What he said lol XD skids
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The radius of the tires of a car is 18 inches, and they are revolving at the rate of 651 revolutions per minute. How fast is the
Mumz [18]
So the car is moving at 651 revolutions per minute, with wheels of a radius of 18inches

so, one revolution, is just one go-around a circle, and thus 2π, 651 revolutions is just 2π * 651, or 1302π, the wheels are moving at that "angular velocity"

now, what's the linear velocity, namely, the arc covered per minute

well   \bf v=rw\qquad 
\begin{cases}
v=\textit{linear velocity}\\
r=radius\\
w=\textit{angular velocity}\\
----------\\
r=18in\\
w=1302\frac{\pi }{min}
\end{cases}\implies v=18in\cdot \cfrac{1302\pi }{min}
\\\\\\
v=\cfrac{23436\pi\ in}{min}

now, how much is that in miles/hrs?  well
let's keep in mind that, there are 12inches in 1foot, and 5280ft in 1mile, whilst 60mins in 1hr

thus   \bf \cfrac{23436\pi\ in}{min}\cdot \cfrac{ft}{12in}\cdot \cfrac{mi}{5280ft}\cdot \cfrac{60min}{hr}\implies \cfrac{23436\cdot \pi \cdot 60\ mi}{12\cdot 5280\ hr}

notice, after all the units cancellations, you're only left with mi/hrs
4 0
2 years ago
How many number of terms does 5p^2-2p+3r have?
Helga [31]

Answer:

3 terms

Step-by-step explanation:

5p^{2} - 2p + 3r

It has 3 terms

They are as follows:

5p^{2}

-2p

and 3r

Hope this helps

plz mark as brainliest!!!!!

7 0
3 years ago
A 250 m train travels through a 2km tunnel at 27 km/hr. How long is it from the time the front of the train enters the tunnel, t
olga55 [171]

Answer:

5 min

Step-by-step explanation:

time taken=distance required to travel/speed of train

total distance train have to cover to just left the tunnel is (lenght of tunnel+length of train) =2km+250m

=2250m

speed of train is 27km/hr

multiply 5/18 to convert in m/s

i.e, (5/18)×27=7.5m/s

now time taken =2250m/(7.5m/s)

=300s

and,300s =5min

4 0
2 years ago
If m∠TRK equal 32.5°, what is m∠TSK?
MrRa [10]
The answer is whatever is equal to 32.5 hope this helped you
7 0
3 years ago
9x^2-9x-18=0 so what will be ans​
dusya [7]

Answer:

9x^2-9x-18=0

9x+2=11x-9x-18=0

9x+2=21x-18=0

9x+2=3x-3^=0

7 0
3 years ago
Read 2 more answers
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