12 divided 2 times 2 ^ 3 equals 120
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Related Questions
The sum of the interior angles of a quadrilateral equals 360°.a. Trueb. False
Computer beeps every 6 mins & chirps every 10 mins, but it beeped & chirped at the same time. in how many minutes will it next make both sounds at the same time
8) A crew is made up of 8 men; the rest are women. 66% of the crew are men. How many people are in the crew?
Rewrite the equation to represent the remaining values on the hanger diagram.
Part 1. Show all work to solve for x in the equation and check the solution.
Part 2. Explain how to identify the extraneous solution and what it means.
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(x + a)/ax = b/x
therefore: a != 0, x !=0
now we solve:
(x + a)/ax = b/x
(x + a)/a = b
x + a = ab
x = a(b - 1)
that is the solution for b, and we keep in mind that a != 0, x !=0, the extraneous solutions
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$0.20 * 523 envelopes = 0.20 * 523 = $104.6
The correct result would be $104.6.
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300 - 0.1n + 0.06n = 270
300 - (0.1n - 0.6n) = 270
300 - 0.04n = 270
-0.04n = -30
n = -30/-0.04 = 750
Investment at 6% = 750
Investment at 10% = 3000 - 750 = 2250
And to check just plug it back in so,
(0.06*750)+(0.1*2250) = 270
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The expression for the total ticket sales if (n) one day passes and (t) three day passes are sold is n + t = $350.
What is an expression?
An expression is a number, or a variable, or a combination of numbers and variables and operation symbols.
Now it is given that,
Cost of one-day passes, n = $100
Cost of three-day passes, t = $250
So, Total ticketssold = Cost of one-day passes + Cost of three-day passes
⇒ Total ticketssold = n + t
⇒ n + t = $100 + $250
⇒ n + t = $350
which is the required expression for total tickets sold.
Thus, the expression for the total ticket sales if (n) one day passes and (t) three day passes are sold is n + t = $350.
To learn more about expressions :
#SPJ2
Answer:
100n+250t
Step-by-step explanation:
cost of the1 one day pass=100
cost of n one day passes=100n
cost of the 1 three day pass=250
cost of t three day passes=250t
Total sale= 100n+250t
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After a night,
f(x,y)=0 ==>df=0==>@f/@x*dx+@f/@y*dy=0
==>dy/dx=-(@f/@x)/(@f/@y) There was a mistake (in my memory)
@f/@x=6xy²+20x^4
@f/@y=5y^4+6x²y
==>dy/dx=-(6xy²+20x^3)/(5y^4+6x²y)
for x=-1 and y=2 dy/dx= -(6*(-1)*4+20*(-1)^3)/(5*2^4+6*(-1)²*2)
=44/92=11/23
or
5y^4*dy/dx+6xy²+6x²ydy/dx+20x^3=0
==>dy/dx*(5y^4+6x²y)=-6xy²-20x^3
==>dy/dx= -(6xy²+20x^3)/(5y^4+6x²y)
==>11/23
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so 4x-7=13
add 7 to both sides
4x=20
divide both sides by 4
x=5
this is one solution
4x-7=-13
add 7 to both sides
4x=-6
divide both sides by 4
x=-1.5
x=5 or -1.5