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Alexxandr [17]
3 years ago
12

Your going 60 miles per hour how long does it take you to reach 180 miles

Mathematics
2 answers:
sveta [45]3 years ago
8 0
It would take you 3 hours, this ja because you go 60 miles per hour and 180/60 = 3
Stells [14]3 years ago
7 0

Answer:

Step-by-step explanation:

The answer is 3 hours.

If you are traveling 60 mph (miles per hour) then in one hour you have traveled 60 miles, in 2 hours 120 miles and finally in 3 hours 180 miles. 60X=180. X=180/60, x=3

Or 60+60+60=180

However you choose to get there it will take 3 hours.

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What is the index of ⁴√81<br>​
ludmilkaskok [199]

Answer:

8

HOPING MAKATULONG PO!,KEEP SMILLING GOD LOVES YOU:)!

3 0
3 years ago
How many terms are in the arithmetic sequence 1313, 1616, 1919, ……, 7070, 7373?
Nataliya [291]
To find the number of terms in the arithmetic sequence, we need to use the formula 
a_{n}= a_{1}+(n-1)d    
where a_{n} is the nth number, a_{1} is the first number, n is the number of terms and d is the difference of the two consecutive numbers.
7373 = 1313 + (n - 1)(303)
7373 = 1313 + 303n - 303
7373 = 1010 + 303n
7373 - 1010 = 303n
6363 = 303n
6363 ÷ 303 = n
n = 21

Therefore, there are 21 terms in the arithmetic sequence given.
7 0
3 years ago
Can somebody help me please.​
Wewaii [24]

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5 0
3 years ago
Read 2 more answers
Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
3 years ago
HELLPPPP! WILL GIVE BRAINLIEST!
MArishka [77]
6.15x = 123
123 ÷ 6.15 = 20
x = 20 hours.
3 0
3 years ago
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