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AveGali [126]
3 years ago
6

the expression (4^p*4^1)^3 simplifies to 4^15. Eric was asked to find the value of p and said the answer is p=11. Is Eric correc

t? why or why not. If not what does p equal?
Mathematics
1 answer:
notka56 [123]3 years ago
8 0

(4^p*4^1)^3  = 4^15

add the exponents

(4^(p+1))^3 = 4^15

power to the power means multiply

(4^3(p+1)) = 4^15

the bases are the same so the powers have to be the same

3(p+1)) = 15

divide by 3

p+1 =5

subtract 1

p=4

Eric is incorrect.  p=4

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Solve without a calculator log_4(20)-log_4(45)+log_4(144)
AfilCa [17]
Log₄20-log₄ 45+log₄144=
log₄(20/45)+log₄144=                        (log_a b- log_a c=log_a  (b/c) )
log₄[(20*144)/45]=                             (log_a b +log_a c=log_a (b*c) )
log₄(2880/45)=
log₄(64)=n     ⇔      4^n=64                (log_a x=n   ⇔ a^n=x)

4^n=4³      ⇒n=3                                 (64=4*4*4=4³)

Answer: log₄20-log₄ 45+log₄144=3

5 0
3 years ago
What is the solution for the system of equations x+y=2 and x-y=4
dmitriy555 [2]
X+y=2 (label equation one) 
x-y=4 (label equation two)

y=-x+2 (equation one rearranged, label this equation equation three)

sub 3 into 2
x-(-x+2)=4
x+x-2=4
2x-2=4
2x=6
x=3
sub x=3 into equation one

3+y=2
y=-1

therefore x=3 and y=-1
5 0
3 years ago
Find ab if the coordinates of a are (3, 5 ( and b are (-3, -3)<br><br>a- 8<br>b- 6<br>c- 12<br>d- 10
strojnjashka [21]
If you use the distance formula you get 10.
6 0
3 years ago
Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t​
juin [17]

Answer:

a. 73; b. 48.9; c. 2; d. 33.8; e. 73

Step-by-step explanation:

Assume the function was

S(t)= 73 - 15 ln(t + 1), t  ≥ 0

a. Average score at t = 0

S(0) = 73 - 15 ln(0 + 1) = 73 - 15 ln(1) = 73 - 15(0) =73 - 0 = 73

b. Average score at t = 4

S(4) = 73 - 15 ln(4 + 1) = 73 - 15 ln(5) = 73 - 15(1.61) =73 - 24.14 = 48.9

c. Average score at t =24

S(24) = 73 - 15 ln(24 + 1) = 73 - 15 ln(25) = 73 - 15(3.22) =73 - 48.28 = 24.7

d. Percent of answers retained

At t = 0. the students retained 73 % of the answers.

At t = 24, they retained 24.7 % of the answers.

\text{Percent retention} = \dfrac{\text{24.7}}{\text{73}} \times 100 \, \% = \text{33.8 \%}\\\\\text{The students retained $\large \boxed{\mathbf{33.8 \, \%}}$ of their original knowledge after two years.}

e. Maximum of the function

The maximum of the function is at t= 0.

Max = 73 %

The graph below shows your knowledge decay curve. Knowledge decays rapidly at first but slows as time goes on.

 

6 0
3 years ago
The value of y varies directly with x. When y = 75, 7 = 1. What is the value of y when x is 272
kow [346]
I’m assuming that 7 means x, and honestly all you have to do is multiply. x= 1 , y= 75.. so just multiply 75 and 272. The answer is 20,400.
8 0
3 years ago
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