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3 years ago

Working out 34 x 17 using an area model

Mathematics

1 answer:

Troyanec [42]3 years ago

4 0

Your answer is going to be 578 because you had to add 300+210+40+28 and it = to 578 your welcome

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Which of the statements is false about the additive identity?

RUDIKE [14]

B) If y is the additive identity is 0

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3 years ago

If e^(xy) = 2, then what is dy/dx at the point (1, ln2)?

love history [14]

E^(xy) = 2
(xdy/dx + y)e^(xy) = 0

At point (1, ln2), dy/dx + ln2 = 0
dy/dx = -ln2

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3 years ago

If g(x)=4x^2-16 were shifted 5 units to the right and 2 down hat would the new equation be

nadezda [96]

Answer:

This is the answer

Step-by-step explanation:

This is the solution, hope it helps

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3 years ago

4. You roll a standard number cube once. Let A

amm1812

Answer:

i thenk it is 12 persent

Step-by-step explanation:

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4 years ago

Plźzzz help me ... I'll give brainliest

fgiga [73]

Answer:

$\displaystyle x = \frac{1}{3}$ .

Step-by-step explanation:

Consider the double angle identity for tangents:

$\displaystyle \tan(2\theta) = \frac{2\tan{\theta}}{1 - \tan^{2}{\theta}}$ .

Also, for two complementary angles,

$\displaystyle \tan{\left(\frac{\pi}{2} - \theta \right)} = \frac{1}{\tan{\theta}}$ .

Subtract $\tan^{-1}(x + 1)$ from both sides of this equation:

$\displaystyle 2\tan^{-1}{x} = \frac{\pi}{2} + (-\tan^{-1}(x + 1))$ .

Take the tangent of both sides of this equation:

$\displaystyle \tan(2\tan^{-1}{x}) = \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)$ .

Apply the double-angle identity to the left-hand side of this equation:

$\displaystyle \tan(2\tan^{-1}{x}) \implies \frac{2\tan(\tan^{-1}x)}{1 - \tan^{2}(\tan^{-1}x)}\implies \frac{2x}{1 - x^{2}}$ .

The two angles $\displaystyle \left(\frac{\pi}{2} + (-\tan^{-1}\left(x + 1\right))\right)$ and $(x - 1)$ are complementary. Therefore, for the right-hand side of this equation,

$\displaystyle \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)\implies \frac{1}{\tan(\tan^{-1}(x + 1))} \implies \frac{1}{x + 1}$ .

Equate the two sides of this equation:

$\begin{aligned} & \frac{2x}{1 - x^{2}} = \frac{1}{x + 1}\\ \implies & \frac{2x}{(1 - x)(1 + x)} = \frac{1}{1 + x}\\\implies & 2x = 1 - x,\; x \ne 1,\; x \ne -1\\\implies & x = \frac{1}{3}\end{aligned}$ .

5 0

3 years ago