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3 years ago

X + |5| = 7 I Don't know how to solve this

Mathematics

1 answer:

max2010maxim [7]3 years ago

4 0

Step-by-step explanation:

x+5=|7

-5 | -5

----------------

x =2

this ^ is how I usually do it

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A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$

when P = 23 kg/cm2, V = 35 cm3, and $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ this becomes

$\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68$

The pressure is changing at $\frac{dP}{dt}=3.68$ .

7 0

3 years ago

A lamina with constant density rho(x, y) = rho occupies the given region. Find the moments of inertia Ix and Iy and the radii of

jenyasd209 [6]

Answer:

Ix = Iy = $\frac{ρπR^{4} }{16}$

Radius of gyration x = y = $\frac{R}{4}$

Step-by-step explanation:

Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.

Mass of disk = ρπR2

Moment of inertia about its perpendicular axis is $\frac{MR^{2} }{2}$ . Moment of inertia of quarter disk about its perpendicular is $\frac{MR^{2} }{8}$ .

Now using perpendicular axis theorem, Ix = Iy = $\frac{MR^{2} }{16}$ = $\frac{ρπR^{4} }{16}$ .

For Radius of gyration K, equate MK2 = MR2/16, K= R/4.

3 0

3 years ago

Write a proportion for the statement. 40 is to 10 as 32 is to 8.

Svetach [21]

Hi! ⋇

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All proportions have this form:

$\sf{\dfrac{a}{b}=\dfrac{c}{d}}$ , Where $\sf{\dfrac{a}{b}}$ is equal to $\sf{\dfrac{c}{d}}$ .

If $\sf{\dfrac{a}{b}\neq\dfrac{c}{d}}$ , it's not a proportion.

_________________________

Here we have two pairs of numbers:

40,10 and 32,8.

Written As a proportion, they look like :

$\sf{\dfrac{40}{10}=\dfrac{32}{8}}$

Hope this made sense to you :)

$\it{Calligrxphy}$

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6 0

1 year ago

If you have 15 blocks and divide the blocks equally among 5 friends, how many blocks does each friend receive?

vaieri [72.5K]

Answer:

Step-by-step explanation:

15 divided by 5 equals 3.

4 0

3 years ago

Read 2 more answers

How do I solve : COT pi/4 ...?

Wewaii [24]

Cot x = cos x / sin x
cot π/4 = cot 45° = cos 45° / sin 45°
We know that sin 45° and sin 45° have the same value:
cos 45° = sin 45° = √2 / 2;
cos 45° / sin 45° = √2/2 : √2/2 = 1
Answer:
cot π/4 = 1

3 0

3 years ago