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d1i1m1o1n [39]
3 years ago
13

If the start time is 8:34 amd the elapsed timne is 3 hours and 18 minutes what is the end time

Mathematics
2 answers:
kompoz [17]3 years ago
4 0
11:52 is the end time.
This is how I solved it. I am not saying that this is the correct way, but it is what it is.
I set up the elapsed time as regular time; hours:minutes>> 3:18
Since the start time is 8:34, and we need to find the ending time, I can simply add the elapsed time and the starting time together.
8:34 +3:18= 11:52

Now, sometimes this won't work. There will be left over time that you'll have to add and everything, but for this situation this way worked.
mr_godi [17]3 years ago
3 0
Forget the '34' part of 8:34 and the '18' minutes from 3 hours and 18 minutes for now.

Just add 3 hours on to 8, to get 11. Then, bring back the 34 and 18 and add them together. 34 + 12 = 52. Now put that 52 on the end of the 11, and you get 11:52 as the end time. 
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3 years ago
if two of these congruent angles have the same degree measure and Angle1 is (3× + 10) degrees and the Angle 2 is (5× - 4) degree
Vitek1552 [10]

Answer:

Each angle = 31 degree

Step-by-step explanation:

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3 years ago
Find the equation of the tangent line to the curve (a lemniscate)
olya-2409 [2.1K]

Answer:

m=\frac{9}{13} and b=\frac{40}{13}

Step-by-step explanation:

The equation of curve is

2(x^2+y^2)^2=25(x^2-y^2)

We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})

4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})

The point of tangency is (-3,1). It means the slope of tangent is \frac{dy}{dx}_{(-3,1)}.

Substitute x=-3 and y=1 in the above equation.

4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})

40(-6+2\frac{dy}{dx})=25(-6-2\frac{dy}{dx})

-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}

80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240

130\frac{dy}{dx}=90

Divide both sides by 130.

\frac{dy}{dx}=\frac{9}{13}

If a line passes through a points (x_1,y_1) with slope m, then the point slope form of the line is

y-y_1=m(x-x_1)

The slope of tangent line is \frac{9}{13} and it passes through the point (-3,1). So, the equation of tangent is

y-1=\frac{9}{13}(x-(-3))

y-1=\frac{9}{13}(x)+\frac{27}{13}

Add 1 on both sides.

y=\frac{9}{13}(x)+\frac{27}{13}+1

y=\frac{9}{13}(x)+\frac{40}{13}

Therefore, m=\frac{9}{13} and b=\frac{40}{13}.

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AnnyKZ [126]

Answer:

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