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3 years ago

My denomination is 6 more than my numerator my simplest form is 2/5 what fraction am i

Mathematics

1 answer:

gladu [14]3 years ago

3 0

Answer:

4/10

Step-by-step explanation:

4/10 is equal to 2/5, and 4+6=10

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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

What is the value of this expression when a = 4, b= -5, and c= -7? a + 2bc 3a ?

sladkih [1.3K]

Answer:

1) -66

Step-by-step explanation:

Given:

a=4

b=-5

c=-7

=a+2bc

=4+2*-5*7

=4+(-10*7)

=4-70

=-66

=3a

=3(4)

=12

Hope this helps ;) ❤❤❤

7 0

3 years ago

Read 2 more answers

At a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective pa

11Alexandr11 [23.1K]

Answer:

Probability that fewer than 2 of these parts are defective is 0.604.

Step-by-step explanation:

We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.

A random sample of 7 parts produced by this machine is chosen.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 7 parts

r = number of success = fewer than 2

p = probability of success which in our question is % of defective

parts produced by one of the machine, i.e; 19%

LET X = Number of parts that are defective

So, it means X ~ Binom(n = 7, p = 0.19)

Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)

P(X < 2) = P(X = 0) + P(X = 1)

= $\binom{7}{0}\times 0.19^{0} \times (1-0.19)^{7-0}+ \binom{7}{1}\times 0.19^{1} \times (1-0.19)^{7-1}$

= $1 \times 1 \times 0.81^{7} +7 \times 01.9^{1} \times 0.81^{6}$

= 0.604

Therefore, the probability that fewer than 2 of these parts are defective is 0.604.

8 0

2 years ago

Find the midpoint of the segment with the given endpoints. (-10.2) and (0,-7)

iris [78.8K]

Answer:

$=\left(-5,\:-\frac{5}{2}\right)$

Step-by-step explanation:

$\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)\\\\\left(x_1,\:y_1\right)=\left(-10,\:2\right),\:\left(x_2,\:y_2\right)=\left(0,\:-7\right)\\\\=\left(\frac{0-10}{2},\:\frac{-7+2}{2}\right)\\\\= (\frac{-10}{2} , \frac{-5}{2})\\ \\=\left(-5,\:-\frac{5}{2}\right)$

7 0

3 years ago

Di ketahui f(x)=ax+b tentukan fungsi f(1)=3danf(2)=5

Komok [63]

F(x) = ax + b
maka..
f(1) = 3
a(1) + b = 3 --> a+b = 3

f(2) = 5
a(2) + b = 5 --> 2a+b = 5

untuk mencari nilai a dan b nya, di eliminasi, maka dapat
a=2 dan b=1

5 0

3 years ago