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2 years ago

Given that A varies directly as B and inversely as C and that; A=12 when B=3 and C=2. Find B when A=10 and C=1.5

Mathematics

1 answer:

maria [59]2 years ago

8 0

Answer:

B = 1.875

Step-by-step explanation:

given that A varies directly as B and inversely as C then the equation relating them is

A = $\frac{kB}{C}$ ← k is the constant of variation

to find k use the condition A = 12 when B = 3 and C = 2 , then

12 = $\frac{3k}{2}$ ( multiply both sides by 2 to clear the fraction )

24 = 3k ( divide both sides by 3 )

8 = k

A = $\frac{8B}{C}$ ← equation of variation

when A = 10 and C = 1.5 , then

10 = $\frac{8B}{1.5}$ ( multiply both sides by 1.5 )

15 = 8B ( divide both sides by 8 )

1.875 = B

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Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

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5 0

2 years ago

1. What is the growth/ decay rate for this exponential function?

Firdavs [7]

Answer:

1. The growth rate is 2

2. The sample is losing half its mass, so the decay factor is 1/2 or 0.5

Step-by-step explanation:

The growth and decay factor in an exponential function is the number that is being multiplied to the power of x

It is a growth rate when that number is greater than 1
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2 years ago

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nataly862011 [7]

Answer:

b. False

Step-by-step explanation:

In a research study, when a researcher wants to find the impact of a new treatment, then the researcher randomly divides the the study participants into two groups. The groups are :

-- control group

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The control group is a group that is used to establish the cause-and-effect relationship by making the effect of an independent variable isolate. It receives no treatment or some standard treatment for the which the effect is already known.

The treatment group receives the treatment for which the effect the researcher is interested in.

Thus the averages of the four categorized groups are not required for estimating the difference.

Therefore, the answer is FALSE.

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3 years ago

Which list shows the lengths from shortest to longest ?

wel

Answer:

It’s a

Step-by-step explanation:

Tell me if I’m wrong

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3 years ago

Please help me on this it’s due

olya-2409 [2.1K]

<h3>Answer: 5 cakes</h3>

================================================

Explanation:

Let's start off converting the mixed number 12 & 1/4 to an improper fraction.

$a \frac{b}{c} = \frac{a*c+b}{c}\\\\12 \frac{1}{4} = \frac{12*4+1}{4}\\\\12 \frac{1}{4} = \frac{49}{4}\\\\$

Do the same for the other mixed number 2 & 1/3.

$a \frac{b}{c} = \frac{a*c+b}{c}\\\\2 \frac{1}{3} = \frac{2*3+1}{3}\\\\2 \frac{1}{3} = \frac{7}{3}\\\\$

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From here, we divide the two fractions. I converted them to improper fractions to make the division process easier.

$\frac{49}{4} \div \frac{7}{3} = \frac{49}{4} \times \frac{3}{7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{49\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 7\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 3}{4}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{21}{4}\\\\$

The last step is to convert that result to a mixed number.

$\frac{21}{4} = \frac{4*5+1}{4}\\\\\frac{21}{4} = \frac{4*5}{4}+\frac{1}{4}\\\\\frac{21}{4} = 4+\frac{1}{4}\\\\\frac{21}{4} = 5 \frac{1}{4}\\\\$

Note that 21/4 = 5.25 and 1/4 = 0.25 to help check the answer.

-----------------------

Therefore, she can make 5 cakes. The fractional portion 1/4 is ignored since we're only considering whole cakes rather than partial ones.

3 0

3 years ago

Given that A varies directly as B and inversely as C and that; A=12 when B=3 and C=2. Find B when A=10 and C=1.5​

Given that A varies directly as B and inversely as C and that; A=12 when B=3 and C=2. Find B when A=10 and C=1.5