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Katena32 [7]
3 years ago
9

Lilli jogs 1/2 mile in 1/10 hour, what is lilli's rate in miles per hour?

Mathematics
2 answers:
joja [24]3 years ago
6 0
If she jogs 1/2 miles in 1/10 hours, we need to find how much she jogs in 1 hour. Since the amount of time (1/10h) went up 10 times (10 * 1/10 = 1) then the distance must’ve also increased 10 times (since time and distance are proportional — the more time she jogs, the more distance she covers) and since 10 times 1/2 is 10/2, which is 5, then we can say she jogs 5 miles per hour.
VikaD [51]3 years ago
3 0

Answer:

5 mph

Step-by-step explanation:


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14x + 64 = <br><br> x = 3<br><br> What is the answer
Nostrana [21]

x = 3.

Plug in 3 for x in the equation

14(3) + 64

Remember to follow PEMDAS. First, multiply 3 with 14

3 x 14 = 42

Finally, add 64

42 + 64 = 106

106 is your answer

hope this helps

5 0
3 years ago
Simplify 4x^2 / 3 X 6/ 8x
Leokris [45]

Answer:

x

Step-by-step explanation:

The answer to

\frac{4 {x}^{2} }{3}  \times  \frac{6}{8x}

= x

3 0
2 years ago
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If you had a 15 foot ladder, how far away from the base of a wall would you have to put it to reach a window 12 feet up?
Kruka [31]

Answer:

9.000000001

Step-by-step explanation:

If the hypotenuse is  15, the ladder, facing the right angle, and the opposite is 12 we need to use SOH to find the angle.

sin of angle is opposite / hypotenuse

sin of angle is 12 / 15

12/15 = 0.8

sin^-1 of 0.8 = 53.13010235

Angle is that

Then we do this again but a different equation

cos of angle is adjacent (x which we are looking for) / hypotenuse

cos of 53.13... is x / 15

(15 x cos of 53.13... )is x

So 15 x cos of 53.13... is 9.000000001 feet away.

5 0
2 years ago
Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec
klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

x=t^3-3t

and

y=t^2-4

The tangent function is given by:

\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }

\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}

The tangent is vertical at when \frac{dx}{dt}=0

\implies \frac{3t^2-3}{2t}=0

\implies 3t^2-3=0

\implies 3t^2=3

\implies t^2=1

\implies t=\pm1

When t=1,

x=1^3-3(1)=-2 and y=1^2-4=-3

When t=-1,

x=(-1)^3-3(-1)=2 and y=(-1)^2-4=-3

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when \frac{dy}{dx}=0 or  \frac{dy}{dt}=0

\implies 2t=0

\implies t=0

When t=0,

x=0^3-3(0)=0 and y=0^2-4=-4

There is a horizontal tangent at (0,-4)

5 0
3 years ago
True or false ? no fake answers please thanks :)
Talja [164]

Answer:

I’m prearranged sure it’s false.

Step-by-step explanation:

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3 years ago
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