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3 years ago

Lucy has a 40-inch T.V. This means the diagonal distance from one corner of the T.V. to the other is 40 inches.

1 answer:

5 0

Using Pathogorean Theorem.....

40 = the square root of 24^2 + 32^2
40 = the square root of 576 + 1024
40 = the square root of 1600
40 = 40

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More comparing decimals

Lisa [10]

When comparing, think of the signs as a crocodile. A crocodile will want to eat greater (bigger) numbers than smaller numbers. So, you can think of it as that way.

2. 0.5 < 0.8 ... 0.8 is greater than 0.5

5. 0.7 = 0.70 ... This is same. 0.70 just has an extra 0 which you don't need.

8. 2.9 > 2.8 ... 2.9 is greater since 9 is bigger than 8.

11. 0.65 > 0.36 ... 65 is greater than 36. Therefore, 0.65 is greater.

14. 4.50 > 4.05 ... 4.50 has a five in the tenth place. 4.05 has five in the hundredth place.

17. 6.01 < 6.1 ... 6.1 has a one in the tenth place. 6.01 has a one in the hundredth place.

5 0

4 years ago

Please help with this

Masja [62]

Its definitely d i did that math

3 0

3 years ago

Melvin drives a delivery truck his company pays him $.48 for every half mile driven and a $35 bonus for every 150 miles driven i

mote1985 [20]

Answer:

The answer to your question is $473.2

Step-by-step explanation:

Data

Payment $0.48 / 0.5 mile

bonus $35 for 150 miles

Distance driven = 420 miles

Process

1.- Calculate the bonus

420 / 150 = 2.8 then he will receive 2 bonuses = 2 x $35

= $70

2.- Calculate the payment for every half mile

420 / 0.5 = 840

840 x 0.48 = $ 403.2

3.- Total payment before taxes

Payment = $403.2 + $70

= $473.2

6 0

4 years ago

Pleasaaaaseee answer the all for BRAINLEST answer and thanks

seropon [69]

For the first one are you looking for g? or the steps

7 0

3 years ago

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