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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

1 answer:

Mandarinka [93]3 years ago

3 0

20 seconds faster than Jesse so one minute and 57 seconds is my best guess.

You might be interested in

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