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3 years ago

Solve for x. logo(x-3) = 2

Mathematics

2 answers:

RUDIKE [14]3 years ago

6 0

Answer:

x=5

Step-by-step explanation:

5-3=2

____ [38]3 years ago

4 0

So the answer is X=5 cs 5-3=2

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The volume of a spherical cancerous tumor is given by the following equation. V(r) =(4/3)pi r^3 If the radius of a tumor is esti

Dimas [21]

Answer:

The error that might occur is $\±0.123cm^3$

Step-by-step explanation:

Given

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

$Radius,\ r = 1.4cm$

$Error = 0.005cm$

Required

Determine the error that might occur in the volume of the tumor

Given that there's an error in measurement, this question will be solved using the concept of differentiation;

First, we'll rewrite the given parameters in differentiation notations;

$r = 1.4$ --- Radius

$dV = \±0.005$ --- Change in measurement

$V(r) = \frac{4}{3}\pi r^3$ --- Volume as a function of radius

The relationship between the above parameters is as follows;

$\frac{dV}{dr} = V'(r)$

This can be rewritten as

$\frac{dV}{dr} = (V(r))'$

Substitute $\frac{4}{3}\pi r^3$ for $V(r)$

$\frac{dV}{dr} = (\frac{4}{3}\pi r^3)'$

Multiply both sides by dr

$dr * \frac{dV}{dr} = (\frac{4}{3}\pi r^3)' * dr$

$dV = (\frac{4}{3}\pi r^3)' * dr$

$dV = \frac{4}{3}\pi( r^3)' dr$

Differentiate

$dV = \frac{4}{3}\pi( r^3)' dr$

$dV = \frac{4}{3}\pi* 3r^2\ dr$

$dV = 4\pi* r^2\ dr$

Substitute the values of r, dr and take $\pi$ as 3.142

$dV = 4 * 3.142* 1.4^2 * \±0.005$

$dV = \±0.1231664$

$dV = \±0.123$ <em>(Approximated)</em>

Hence, the error that might occur is ±0.123

6 0

3 years ago

A wave is modeled with the function y = 1/2 sin 3 theta. Describe the graph of this function, including its period, amplitude, a

PolarNik [594]

It is a sinusoidal graph, which oscilates around the x-axis (y=0), from y = -0.5 to y = 0.5

General description (as reference)
It passes through the origin (0,0), increases until the max value (y = 0.5) when x = π6, then start to decrease, crosses the x-axis x = π/3, continues decreasing until the min value (y = -0.5) when x = 2π/3, then it starts to increase, crossing the x-axis again at x = 2π/3, and the pattern repeats indefinetely to the right of x = 2π/3 and to the left of x = 0.

Specific description:

Amplitud: 0.5----> the coefficient in front of the sine function

Period: 2π/3 ----> 2π divided by the coefficient in front of the argument (3)

Points of intersection with the x-axis: 0 + /- nπ/3, with n = 0,1,2,3,.....

Type of function: odd

Maximun value: 1/2

Minimum value: -1/2

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4 years ago

Write an equation of the line passing through the points (−5, 6), (−7, 3)

blsea [12.9K]

Answer:

all work is shown and pictured