Answer: Choice D
b greater-than 3 and StartFraction 2 over 15 EndFraction
In other words,
b > 3 & 2/15
or
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Explanation:
Let's convert the mixed number 2 & 3/5 into an improper fraction.
We'll use the rule
a & b/c = (a*c + b)/c
In this case, a = 2, b = 3, c = 5
So,
a & b/c = (a*c + b)/c
2 & 3/5 = (2*5 + 3)/5
2 & 3/5 = (10 + 3)/5
2 & 3/5 = 13/5
The inequality 
is the same as 
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Let's multiply both sides by 15 to clear out the fractions
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Now isolate the variable b
Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how
47/15 = 3 remainder 2
The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.
An exponential function can be expressed as:
y=ab^x we have two points so we can solve
64/8=(ab^4)/(ab^1)
8=b^3
2=b, now we have:
y=a(2^x), using point (4,64)
64=a(2^4)
64=16a
4=a so
y=4(2^x)
Answer:
22
Step-by-step explanation:
Pretend the 10 values in the first sentence are a,b,c,d,e,f,g,h,i,j
Pretend the addition 5 values is k,l,m,n,o
So the mean of all the 15 data is (a+b+c+d+e+f+g+h+i+j+k+l+m+n+o)/15=20
So the sum of all 15 data is a+b+c+d+e+f+g+h+i+j+k+l+m+n+o=300 since 15(20)=300
Now let's look at the first 10: We have their mean so we can write:
(a+b+c+d+e+f+g+h+i+j)/10=19
so a+b+c+d+e+f+g+h+i+j=190 since 10(19)=190
So that means using our first sum equation and our equation sum equation we have
190+k+l+m+n+o=300
k+l+m+n+o=300-190
k+l+m+n+o= 110
So the average of those 5 numbers mentioned in your problem is 110/5=22
Answer:
2x+3
Step-by-step explanation:
the upper left, upper right, and lower right are all using distributive property. The lower left is not.
Answer:
answer given in step by step...
Step-by-step explanation:
13x+13(x+1)+13(x+2)=312
13x+13x+13+13x+26=312
39x+39=312
39x=312-39
39x=273
x=7
first number=13x=13*7=91
second number=13(8)=104
third number=13(9)=117
