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3 years ago

What is the value of 7x-2x+3,when x=2

2 answers:

8 0

Well if x = 2, then you do 7 times 2, which is 14, then 2 times 2 which is 4, then 14 minus 4 is 10, and 10 plus 3 is 13.

ollegr [7]3 years ago

5 0

BIDMAS

2x = 4
4 + 3 = 7
7x = 14
14 - 7 = 7

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Assume you have such a watch. if a minimum of 18.0% of the original tritium is needed to read the dial in dark places, for how m

laila [671]

Given that the half life of 3H is 12.3 years.

The amount of substance left of a radioactive substance with half life of $t_{ \frac{1}{2} }$ after t years is given by

$N(t)=N_0\left(\frac{1}{2} \right)^{ \frac{t}{t_{\frac{1}{2}} }$

Therefore, the number of years it will take for 18% of the original tritinum to remain is given by

$18 \% = 100 \% \left(\frac{1}{2} \right)^{ \frac{t}{12.3}} \\ \\ \Rightarrow\left(\frac{1}{2} \right)^{ \frac{t}{12.3}} =0.18 \\ \\ \Rightarrow\frac{t}{12.3}\ln\left(\frac{1}{2} \right)=\ln0.18 \\ \\ \Rightarrow\frac{t}{12.3}= \frac{\ln0.18}{\ln\left(\frac{1}{2} \right)} = \frac{-1.715}{-0.6931} =2.474\\ \\ \Rightarrow t=2.474(12.3)=30.4$

Therefore, the number of <span>years that the time could be read at night is 30.4 years.</span>

5 0

3 years ago

If $396 is invested at an interest rate of 13% per year and is compounded continuously, how much will the investment be worth in

Vilka [71]

Answer: First option is correct.

Step-by-step explanation:

Since we have given that

Amount invested = $396

Interest rate = 13% per year

Time = 3 years

We will use the "Exponential growth formula" i.e.

$A=Pe^{rt}$

Here,

P denotes Principle amount

r denoted rate of growth

t denotes number of years,

So, it becomes

$A=396e^{0.13\times 3}\\\\A=396e^{0.69}\\\\A=\$584.88$

Hence, First option is correct.

3 0

3 years ago

Read 2 more answers

I’m just lost on this on

Nikitich [7]

Answer:

im sorry i cant really see that...

Step-by-step explanation:

5 0

2 years ago

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = yzi + 4xzj + ex

natima [27]

Answer:

The result of the integral is 81π

Step-by-step explanation:

We can use Stoke's Theorem to evaluate the given integral, thus we can write first the theorem:

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S$

Finding the curl of F.

Given $F(x,y,z) = < yz, 4xz, e^{xy} >$ we have:

$curl \vec F =\left|\begin{array}{ccc} \hat i &\hat j&\hat k\\ \cfrac{\partial}{\partial x}& \cfrac{\partial}{\partial y}&\cfrac{\partial}{\partial z}\\yz&4xz&e^{xy}\end{array}\right|$

Working with the determinant we get

$curl \vec F = \left( \cfrac{\partial}{\partial y}e^{xy}-\cfrac{\partial}{\partial z}4xz\right) \hat i -\left(\cfrac{\partial}{\partial x}e^{xy}-\cfrac{\partial}{\partial z}yz \right) \hat j + \left(\cfrac{\partial}{\partial x} 4xz-\cfrac{\partial}{\partial y}yz \right) \hat k$

Working with the partial derivatives

$curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(4z-z\right) \hat k\\curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k$

Integrating using Stokes' Theorem

Now that we have the curl we can proceed integrating

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S$

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot \hat n dS$

where the normal to the circle is just $\hat n= \hat k$ since the normal is perpendicular to it, so we get

$\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S \left(\left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k\right) \cdot \hat k dS$