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Molodets [167]
3 years ago
8

Write an expression to represent: Nine less than the quotient of two and a number x

Mathematics
1 answer:
Sphinxa [80]3 years ago
5 0

Answer:

2÷x-9

Step-by-step explanation:

You just have to try to find out what each part means.

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On a coordinate plane, triangle A B C has points (4, 4), (7, 8), (10, 4). ... Hence, the scale factor is 2 and one-half( ).

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The center of dilation is a fixed point in the plane. If the scale factor is greater than 1, the image is an enlargement (a stretch). If the scale factor is between 0 and 1, the image is a reduction (a shrink). If the scale factor is 1, the figure and the image are congruent.

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Solve for x 2/3 + 3/5x = x+5/10<br> x = –3<br> x = 2<br> x = –3 and x = 2<br> No Solution
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c

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4 0
3 years ago
Find dy/dx of the function y = √x sec*-1 (√x)​
ELEN [110]

Hi there!

\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) +  \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}

y = \sqrt{x} * sec^{-1}(-\sqrt{x}})

Use the chain rule and multiplication rules to solve:

g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

g(f(x)) = g'(f(x)) * 'f(x))

Thus:

f(x) = √x

g(x) = sec⁻¹ (√x)

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}

Simplify:

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) +  \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}

5 0
3 years ago
Read 2 more answers
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