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Alja [10]
3 years ago
14

Can you show me how you worked this: Evaluate the expression. 9! − 7!

Mathematics
1 answer:
Aloiza [94]3 years ago
3 0
I hope this helps you



9.8.7!-7!


7! (72-1)

7!.71
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I keep trying to use the Pythagorean Theorem to find the missing sides length but it comes to a -31 then an Error when square ro
OverLord2011 [107]
Diagonal of the left triangle = ((8)^2+(7)^2)^0.5= (113)^0.5
x =( 12^2 -(113)^0.5)^2))^0.5=( 144- 113)^0.5= (31)^0.5= 5.57


For ur working, ur mistake lies in the line : x^2+12^2=10.6^2
It shld be 12^2= x^2+10.6^2
3 0
4 years ago
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A particle moves on the hyperbola xy=18 for time t≥0 seconds. At a certain instant, y=6 and dydt=8. What is x that this instant?
professor190 [17]

Answer:

The value of x at this instant is 3.

Step-by-step explanation:

Let x\cdot y = 18, we get an additional equation by implicit differentiation:

x\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0 (1)

From the first equation we find that:

x = \frac{18}{y} (2)

By applying (2) in (1), we get the resulting expression:

\frac{18}{y}\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0 (3)

y\cdot \frac{dx}{dt}=-\frac{18}{y}\cdot \frac{dy}{dt}

\frac{dx}{dt} = -\frac{18}{y^{2}} \cdot \frac{dy}{dt}

If we know that y = 6 and \frac{dy}{dt} = 8, then the first derivative of x in time is:

\frac{dx}{dt} = -\frac{18}{6^{2}} \cdot (8)

\frac{dx}{dt} = -4

From (1) we determine the value of x at this instant:

x\cdot \frac{dy}{dt} = -y\cdot \frac{dx}{dt}

x = -y\cdot \left(\frac{\frac{dx}{dt} }{\frac{dy}{dt} } \right)

x = -6\cdot \left(\frac{-4}{8} \right)

x = 3

The value of x at this instant is 3.

4 0
3 years ago
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natta225 [31]

Answer:

It seems that problem might have an error.

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day 2 population = 600 * 2.718281828459 ^(.02*2) = 88.67

We can see that the population doubles in less than one day  but the lowest of those multiple choices is 34.657 days

Step-by-step explanation:

6 0
3 years ago
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MArishka [77]

Answer: B

Step-by-step explanation: 3^4 is 81, and 3^2 is 9. 81/9 is 9. :)

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3 years ago
A polynomial has been factored below, but some constants are missing. 2x^3 –6x^2 –8x=ax(x+b)(x+c) What are the missing values of
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2x (x - 4) (x + 1). A=2, B=-4, C=1
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