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3 years ago

Order 34 × 102, 1.2 × 107, 8.11 × 10–3 and 435 from least to greatest.

1 answer:

andrey2020 [161]3 years ago

6 0

I will assume that this question is about scientific notation (also assuming that the first number is 3.4*10^2), where you can arrange the numbers by the exponent of the 10, so from least to greatest:
$8.11*10^{-3}$ , $3.4*10^{2}$ , 435, and $1.2*10^{7}$

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Claire said that 10/15 of her CDs are country music. Which statement could be true based upon this relationship?

never [62]

Answer:

C. Claire has 10 country CDs and 5 Jazz CDs

Step-by-step explanation:

$\dfrac{10}{15}$ means that 10 out of 15. So we can assume that in total she has 15 CDs and 10 of them are Country and the other is Jazz.

This would rule out choice D.

If we reduce this fraction then we would have:

$\dfrac{10}{15} = \dfrac{2}{3}$

This means that two-thirds of her CDs are country music. So this would rule out A and B choices.

Now C is correct because 10 country plus 5 Jazz is equal to 15 CDs. So we have the correct total.

The answer is C.

7 0

3 years ago

Read 2 more answers

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:

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

5 0

2 years ago

How can I use multiplication to find the quotient and evaluate the expression

adell [148]

When you divide two fractions, you do Keep, Change, Flip or KCF. So you keep the 3/5, you change the division sign to a multiplication sign, then flip the 3/15 to 15/3. So the new equation is 3/5x15/3. Then you simply the two sides so it becomes 1/1x3/1.....

Meaning the answer is 3.

8 0

3 years ago

There is 2 questions. PLEASE HELP ME

attashe74 [19]

B and c I believe ..........

4 0

2 years ago

When finding the length of the hypotenuse of a right triangle why might you approximate the answer?

sammy [17]

Answer:

Because the hypotenuse might not be a perfect square

Step-by-step explanation:

8 0

2 years ago