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Elenna [48]
3 years ago
12

Evaluate 2y when y = 6

Mathematics
1 answer:
Alina [70]3 years ago
4 0

\bf \red{answer}

When,

y=6

2y=2×6=12

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X ⋅ x = ? (1 point) jjjj
liraira [26]

Answer:

x-x = 0

x+x= 2x

x×x= x^2

have a nice day...

4 0
2 years ago
Wilhelmina wrote a check for $72 to pay her monthly Internet bill, but when
lesantik [10]

Answer:

her bank statement will show that she has $144 less than her checkbook balance

Step-by-step explanation:

let the amount of money in the bank before writing the check be x

after writing the check for $72, because she accidentally recorded this as a credit, here checkbook will reflect a balance of :$(x + 72)

However, the bank correctly records the debit of the $72, resulting  a balance of: $(x-72)

Hence compared to her checkbook,  the bank records will show an amount difference of :

(x-72) - (x + 72)

= x - 72 -x -72

= -144

i.e her bank statement will show that she has $144 less than her checkbook balance

3 0
3 years ago
Read 2 more answers
Given a right cone with base area b and height h, what is the formula for the volume?
olga nikolaevna [1]
The formula for a right cone:

V = πr^2(h/3)

V: Volume
r: radius
h: height
πr^2 = base area of a cone

Answer
The formula for the volume in terms of base area b and height h is: V = b(h/3)
3 0
3 years ago
Read 2 more answers
Find the value of x.<br><br> log 3 x = 4<br><br> A.) 7<br> B.) 12<br> C.) 27<br> D.) 81
lakkis [162]

Hello from MrBillDoesMath!

Answer:

Choice D, 81

Discussion:

By log 3x = 4 I presume that "3" is the base not a multiplier of x. Assuming it is the base, then the equation is equivalent to

x = 3^4

  = 81

This is Choice D.


Thank you,

MrB

8 0
3 years ago
Solve the decay equation y' = -ky using the integrating factors method. Show work
Alex777 [14]

Answer:

y=\frac{c}{e^{kx}}

Step-by-step explanation:

\frac{\mathrm{d} y}{\mathrm{d} x} =-ky

\frac{\mathrm{d} y}{\mathrm{d} x} +ky=0

comparing with equation

\frac{\mathrm{d} y}{\mathrm{d} x} + Py=Q(x)

I.F.= e^{\int P dx}

I.F.= e^{\int k dx}

I.F.= e^{kx}

y=\frac{1}{I.F.} ( \int {Q(x)}  dx  +c)

y=\frac{1}{e^{kx}} ( \int {0}  dx  +c)

y=\frac{c}{e^{kx}}

7 0
3 years ago
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