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garri49 [273]
3 years ago
14

A cyclist rode the first 30-mile portion of his workout at a constant speed. For the 24-mile cooldown portion of his workout, he

reduced his speed by 3 miles per hour. Each portion of the workout took the same time. Find the cyclist's
speed during the first portion and find his speed during the cooldown portion
The cyclist's speed during the first portion is miles per hour.
(Simplify your answer.)
Mathematics
1 answer:
LuckyWell [14K]3 years ago
3 0

what math is this could you tell me what math this is plz

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Given w = 1 + i, what is arg (w) and |w|? Explain how you got your answer.
kirill [66]

Answer:

arg (w)=π/4 and  |w|=√2

Step-by-step explanation:

If z= x+iy is a complex number, then |z| is given by √ ( x²+y² ).

And the argument (z) is given by Ф = tan⁻¹ ( y/x ) where -π < x < π.

Now, in our case w= 1+i i.e. x=1 and y=1.

Therefore, arg (w) = tan⁻¹ (1/1) = π/4 and |w|=√ ( 1²+1² ) = √2 (Answer)

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3 years ago
What is mathematical model
serg [7]
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3 0
3 years ago
Read 2 more answers
Solve the initial value problem
tigry1 [53]

9(t+1)\dfrac{\mathrm dy}{\mathrm dt}-7y=14t\implies\dfrac{\mathrm dy}{\mathrm dt}-\dfrac7{9(t+1)}y=\dfrac{14t}{9(t+1)}

Look for an integrating factor \mu(t):

\ln\mu=\displaystyle-\frac79\int\frac{\mathrm dt}{t+1}=-\frac79\ln(t+1)\implies\mu=(t+1)^{-7/9}

Multiply both sides by \mu:

(t+1)^{-7/9}\dfrac{\mathrm dy}{\mathrm dt}-\dfrac79(t+1)^{-16/9}y=\dfrac{14}9t(t+1)^{-16/9}

Condense the left side as the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[(t+1)^{-7/9}y\right]=\dfrac{14}9t(t+1)^{-16/9}

Integrate both sides:

(t+1)^{-7/9}y=\displaystyle\frac{14}9\int t(t+1)^{-16/9}\,\mathrm dt

For the integral on the right, substitute

u=t+1\implies t=u-1\implies\mathrm dt=\mathrm du

\displaystyle\int t(t+1)^{-16/9}\,\mathrm dt=\int(u-1)u^{-16/9}\,\mathrm du

\displaystyle=\int\left(u^{-7/9}-u^{-16/9}\right)\,\mathrm du=\frac92u^{2/9}+\frac97u^{-7/9}+C

\implies(t+1)^{-7/9}y=\dfrac{14}9\left(\dfrac92(t+1)^{2/9}+\dfrac97(t+1)^{-7/9}+C\right)

\implies(t+1)^{-7/9}y=7(t+1)^{2/9}+2(t+1)^{-7/9}+C

\implies y=7(t+1)+2+C(t+1)^{7/9}=7t+9+C(t+1)^{7/9}

Given that y(0)=12, we get

12=9+C\implies C=3

\implies\boxed{y(t)=7t+9+3(t+1)^{7/9}}

6 0
3 years ago
A box contains 10 colored balls. 5 balls are green, 3 balls are red and 2 balls are yellow. One ball is selected at random. What
Setler79 [48]
There is more green balls so that would probably be picked 
6 0
4 years ago
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