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2 years ago

That’s my question ^^

Mathematics

2 answers:

Blababa [14]2 years ago

6 0

Answer:

A. is the answer hop was this helps I did that problem

beks73 [17]2 years ago

5 0

The answer to the question is A.

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How do you solve this question? Any help appreciated!

fiasKO [112]

$\bf \displaystyle \int\limits_{1}^{e}\cfrac{1}{t}\cdot dt\\\\
-------------------------------\\\\
\textit{doing substitution}\\\\
u=\cfrac{1}{t}\implies u=t^{-1}\implies \cfrac{du}{dt}=-t^{-2}\implies \cfrac{du}{dt}=-\cfrac{1}{t^2}\\\\\\ -t^2du=dt\\\\
-------------------------------\\\\
\displaystyle \int\limits_{1}^{e}u\cdot -t^2du\impliedby \textit{now, let's do some substitution on the "t"}\\\\
-------------------------------\\\\$

$\bf u=\cfrac{1}{t}\implies t=\cfrac{1}{u}\implies t^2=\cfrac{1^2}{u^2}\implies t^2=\cfrac{1}{u^2}\\\\
-------------------------------\\\\
\displaystyle \int\limits_{1}^{e}u\cdot -\cfrac{1}{u^2}\cdot du\implies -1\int\limits_{1}^{e}\cfrac{1}{u}\cdot du\implies \left. -ln|u| \cfrac{}{}\right]_1^e$

$\bf \left. -ln\left( \frac{1}{t} \right) \cfrac{}{}\right]_1^e\implies 
\left[ -ln\left( \frac{1}{e} \right) \right]-\left[ -ln\left( \frac{1}{1} \right) \right]\implies 
\left[ -ln\left( e^{-1}\right) \right]-\left[ -ln\left( 1\right) \right]
\\\\\\\
[-(-1)]-[-(0)]\implies 1-0\implies 1$

8 0

2 years ago

Solve the system.

anyanavicka [17]

X= -2
Y= -1
Z= 1
You may use solve this problem by substituting the values as well. If everything goes wrong , go for Cramer’s rule

3 0

3 years ago

If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.

zmey [24]

Answer:

$a=3$ and $b=5$ .

Step-by-step explanation:

So I believe the problem is this:

$\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}$

where we are asked to find values for $a$ and $b$ such that the equation holds for any $x$ in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

$\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)$

$2(4x+3)=a(x+7)+b(x-3)$

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

$2(4\cdot -7+3)=a(-7+7)+b(-7-3)$

$2(-28+3)=a(0)+b(-10)$

$2(-25)=0-10b$

$-50=-10b$

Divide both sides by -10:

$\frac{-50}{-10}=b$

$5=b$

Now we have:

$2(4x+3)=a(x+7)+b(x-3)$ with $b=5$

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

$2(4\cdot 3+3)=a(3+7)+b(3-3)$

$2(12+3)=a(10)+b(0)$

$2(15)=10a+0$

$30=10a$

Divide both sides by 10:

$\frac{30}{10}=a$

$3=a$

So $a=3$ and $b=5$ .

4 0

2 years ago

Read 2 more answers

Two bags contain coloured cubes.

spayn [35]

Answer:

equal

Step-by-step explanation:

Bag 1 has 3 red cubes and 1 blue cube. = 4 cubes

Bag 2 has 3 green cubes, 1 yellow cube, and 1 red cube.= 5 cubes

P ( red bag 1) = red /total = 3/4

P (yellow bag 2) = yellow /total = 1/5

P(red bag1 , yellow bag2) = 3/4* 1/5 = 3/20

P ( blue bag 1) =blue /total =1/4

P (green bag 2) = green /total = 3/5

P(blue bag1 , green bag2) = 1/4* 3/5 = 3/20

The probabilities are equal

8 0

3 years ago

Partial fraction decomposition<br> 20x+9/25x^2+20x+4

ozzi

Step-by-step explanation:

We have

$\frac{20x + 9}{25 {x}^{2} + 20x + 4 }$

Let's factor the denomiator first,

the denomaitor is a perfect square so we get

$\frac{20x + 9}{(5x + 2) {}^{2} }$

Now, we must think of two fractions that

$\frac{a}{(5x + 2) {}^{2} } + \frac{b}{5x + 2}$

We use a perfect square term for one fraction, then a linear one for the next, because if we set both of the denomiator to the same factor, we would get a inconsistent system.

So right now, we have

$\frac{a}{(5x + 2) { }^{2} } + \frac{b}{5x + 2} = \frac{20x +9 }{25 {x}^{2} + 20x + 4 }$