Which graph shows the solution to the inequality -0.5x s 7.5?

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

3 0

Answer:

your answer is for sure please see the solution

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How many almonds are in the bag?

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Almonds in hypothetical bag of which we cannot see

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3 years ago

Using the ration of perfect squares method, what is the square root of 96 rounded to the nearest tenth

laila [671]

You might need to consult your teacher or text to learn the details of the "ratio of perfect squares method" of determining an approximate square root. No reference to such a method can be found in an Internet search except in conjunction with problems similar to this one.

A method that can be used to find a first approximation of a square root is linear interpolation between the roots of adjacent perfect squares. For this, ...
• find the perfect squares of the consecutive integers that lie on either side of the root of interest
• form the ratio of difference between the number and the smaller square and the difference between the squares
• add this ratio to the smaller of the two integers to get an approximation of the root.

In this case, the square root of 96 lies between 9² = 81 and 10² = 100. The ratio of interest is (96 -81)/(100 -81) = 15/19. The approximate square root of 96 is then ...
√96 ≈ 9 + 15/19 ≈ 9.8

_____
If a = floor(√n), then this approximation to the root can be written as
√n ≈ a + (n -a²)/(2a+1)

If we define b = n - a², this looks like √n ≈ a + b/(2a+1). The approximation can be refined by replacing the 1 in the denominator with (b/(2a+1)). Repeately doing this replacement results in a "continued fraction" that converges to √n as more layers are added.

6 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:

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