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3 years ago

7

Pre-calc, Review the table of values for function g(x). Which statement correctly explains whether Limit of g (x) as x approache

s 3 exists? (image attached)

1 answer:

olganol [36]3 years ago

7 0

Answer:

$\lim_{x \to 3} g(x) = DNE$

Step-by-step explanation:

Observations:

As x approaches 3 from the left side, f(x) approaches -1.6
As x approaches 3 from the right side, f(x) approaches 1.6

Conclusion:

Because the right and left-side limits are not equal to each other, the limit does not exist
Therefore, $\lim_{x \to 3} g(x) = DNE$

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Answer:

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Step-by-step explanation:

The distance between two points A(x₁, y₁) and B(x₂, y₂) on the coordinate plane is given by:

$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\$

Point J is at (-3, 3), point K is at (4, 3) and point L is at (1, -1). Hence:

The distance between K and L = KL = $\sqrt{(-1-3)^2+(1-4)^2} =5\ units$

The distance between J and L = JL = $\sqrt{(-1-3)^2+(1-(-3))^2} =4\sqrt{2} \ units$

The distance between K and J = JK = $\sqrt{(3-3)^2+(-3-4)^2} =7\ units$

Therefore, the perimeter of triangle JKL is:

Perimeter = KL + JL + JK = 5 + 4√2 + 7 = 17.66 units

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3 years ago

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Read 2 more answers

A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a

miv72 [106K]

Answer:

22670.8 cm³/min

Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, $\frac{dr}{dt}=5\ cm/min$

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

So, first we find the volume of the sphere in terms of 'r'. As the balloon is spherical in shape, the volume of the balloon is equal to the volume of a sphere. Therefore,

Volume of balloon is given as:

$V=\frac{4}{3}\pi r^3$

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

$\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})$

Now, plug in 19 cm for 'r', 5 cm per minute for $\frac{dr}{dt}$ and solve for $\frac{dV}{dt}$ . This gives,

$\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min$

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

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4 years ago