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2 years ago

Is p(x)= |x| an inverse function

Mathematics

2 answers:

Romashka-Z-Leto [24]2 years ago

7 0

G(x) = |xl would be the inverse baby

docker41 [41]2 years ago

4 0

Answer:

Step-by-step explanation:

p(x) = |x|

in : R let : y = |x|

calculate : x

y² = |x| ²

y² = x²

y² - x²= ( y - x)( y +x) by identity

y = x or y = - x

the inverse : y = x or y = - x

conclusion : the inverse is : g(x) = |x|

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Andru [333]

Answer:A,B,E

Step-by-step explanation:

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3 years ago

a 9ft ladder is placed 4.5 feet from the base of a wall. How high up will the ladder reach? Round to the nearest hundredth.

spin [16.1K]

Answer:

$\huge\boxed{7.79 \ \text{feet}}$

Step-by-step explanation:

We know that we have a 9 foot long ladder resting upon a wall, with an unknown height, but the ladder is 4.5 feet away from the wall.

This can be represented as a triangle, where x is the unknown side.

| \

x | \ 9

| \

|_____\

4.5

We can use The Pythagorean Theorem to find the value of x. The theorem states that $a^2+b^2=c^2$ , where a and b are the lengths of the legs and c is the length of the hypotenuse.

However, we already know the hypotenuse and one leg. Therefore we can substitute inside the equation to find the missing value.

$a^2 + 4.5^2 = 9^2\\\\a^2 + 20.25 = 81\\\\a^2 = 60.75\\\\a \approx 7.79$

Hope this helped!

7 0

2 years ago

Chang knows one side of a triangle is 13 cm. Which set of two sides is possible for the lengths of the other two sides of this t

Lelechka [254]

8cm and 8cm.

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2 years ago

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.