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nlexa [21]
3 years ago
15

Jessie finish the first race in two minutes and seven seconds 10 finish 12 seconds faster than Jesse one finished eight seconds

faster than 10 how long did it take one to finish
Mathematics
1 answer:
Mandarinka [93]3 years ago
3 0
20 seconds faster than Jesse so one minute and 57 seconds is my best guess.
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Which lists the ratios in order from least to greatest 2:9,1:6,7:36
mezya [45]

Answer:

1:6, 7:36, 2:9

Step-by-step explanation:

2 : 9 → 8 : 36

1 : 6 → 6 : 36

7 : 36

Least → Greatest

1:6, 7:36, 2:9

4 0
3 years ago
What is the logarithmic form of the solution to 102t = 9?
Harman [31]

Equivalent equations are equations that have the same value

The equation in logarithmic form is t = \frac{\log(9)}{2}

<h3>How to rewrite the equation</h3>

The expression is given as:

10^{2t} = 9

Take the logarithm of both sides

\log(10^{2t}) = \log(9)

Apply the power rule of logarithm

2t\log(10) = \log(9)

Divide both sides by log(10)

2t = \frac{\log(9)}{\log(10)}

Apply change of base rule

2t = \log_{10}(9)

Divide both sides by 2

t = \frac{\log_{10}(9)}{2}

Rewrite as:

t = \frac{\log(9)}{2}

Hence, the equation in logarithmic form is t = \frac{\log(9)}{2}

Read more about logarithms at:

brainly.com/question/25710806

8 0
2 years ago
Plz help thankssssssssssssssss
Kisachek [45]

what do you need help with

5 0
3 years ago
Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).
lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted C, which we can parameterize by the vector-valued function,

\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)

for 0\le t\le1, which has differential

\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

Then with x(t)=y(t)=z(t)=1-t, we have

\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r

=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt

Complete the square in the quadratic term of the integrand: t^2-2t+2=(t-1)^2+1=(1-t)^2+1, then in the integral we substitute u=1-t:

\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt

\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du

Make another substitution of v=u^2:

\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv

Integrate by parts, taking

r=v+1\implies\mathrm dr=\mathrm dv

\mathrm ds=e^v\,\mathrm dv\implies s=e^v

\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv

\displaystyle=-(2e-1)+(e-1)=-e

So, we have by the fundamental theorem of calculus that

\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)

\implies-e=f(1,1,1)-2

\implies f(1,1,1)=2-e

3 0
3 years ago
39 s 150% of what number?
professor190 [17]

Answer:

26

Hope this helps!

6 0
2 years ago
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