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2 years ago

30+ Points!!! 5. Solve the following inequalities. a) 2 log3x – 2 logx3 -3 <0

Mathematics

1 answer:

Ilya [14]2 years ago

6 0

Answer:

I answered your last question also

2 log3x – 2 logx3 -3 <0

$\mathrm{Subtract\:}2\log ^3\left(x\right)\mathrm{\:from\:both\:sides}$

$2\log ^3\left(x\right)-2logx^3-3-2\log ^3\left(x\right)$

$\mathrm{Simplify}$

$-2logx^3-3$

$\mathrm{Add\:}3\mathrm{\:to\:both\:sides}$

$-2logx^3-3+3$

$\mathrm{Simplify}$

$-2logx^3$

$Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)$

$\left(-2logx^3\right)\left(-1\right)>-2\log ^3\left(x\right)\left(-1\right)+3\left(-1\right)$

$\mathrm{Simplify}$

$2lx^3og>2\log ^3\left(x\right)-3$

$\mathrm{Divide\:both\:sides\:by\:}2lx^3o;\quad \:l>0$

$\frac{2lx^3og}{2lx^3o}>\frac{2\log ^3\left(x\right)}{2lx^3o}-\frac{3}{2lx^3o};\quad \:l>0\\$

$\mathrm{Simplify}$

$g>\frac{2\log ^3\left(x\right)-3}{2lx^3o};\quad \:l>0$

Step-by-step explanation:

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(x2 – 3x + 2)/(4x - 16)

Tom [10]

Answer:

Pretty sure it's x^2−3x+2/4x-16

Step-by-step explanation:

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2 years ago

Find the equation of the line passing through the points (2, 4) and (3, 2). y = [? ]x + [ ]

PtichkaEL [24]

Answer:

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m=(4-2)/(2-3)

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y-y1=m(x-x1)

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2 years ago

Solve 13x - 4 < 12(x - 1).

Effectus [21]

13x - 4 < 12(x-1)
13x - 4 < 12x - 12
13x - 12x < -12 + 4
x < -8
x > 8

x is greater than 8

6 0

2 years ago

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min. The volume of a sphere is V = 4 3 πr3 and its surface

Scrat [10]

Answer:

The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.

Step-by-step explanation:

The area of a sphere is given by the following formula:

$A = 4\pi r^{2}$

In which A is the area, measured in cm², and r is the radius, measured in cm.

Assume that the radius r of a sphere is expanding at a rate of 40 cm/min.

This means that $\frac{dr}{dt} = 40$

Determine the rate of change in surface area when r = 20 cm.

This is $\frac{dA}{dt}$ when $r = 20$ . So

$A = 4\pi r^{2}$

Applying implicit differentiation.

We have two variables, A and r, so:

$\frac{dA}{dt} = 8r\pi \frac{dr}{dt}$

$\frac{dA}{dt} = 8*20\pi*40$

$\frac{dA}{dt} = 20106.19$

The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.

6 0

2 years ago

Need answers ASSAP

ollegr [7]

Data:
r (radius) = 3.5 in
h (height) = 8 in
Adopt: $\pi \approx3.14$
V (volume) = ?
Ink cost per cubic inch = $0.05/in³
Total cost: ?

Calculate the volume of the paint bucket (its formed being cylindrical), we have:
 $V = \pi *r^2*h$
$V = 3.14 *3.5^2*8$
$V = 3.14*12.25*8$
$\boxed{V = 307.72\:in^3}$

What was the total cost of the paint?

price($)___measure(in³)
0.05 ---------- 1
y -------------- 307.72

It is made the rule of three: (directly proportional)

$\frac{0.05}{y} = \frac{1}{307.72}$
multiply cross
$1*y = 0.05*307.72$
$y = 15.386\:\to\:\boxed{\boxed{y\approx\$15}}\end{array}}\qquad\quad\checkmark$

Answer:
the total cost of the paint is $15

7 0

3 years ago