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Yuki888 [10]
3 years ago
15

Tiles. Show your work5. X+4= 4​

Mathematics
1 answer:
omeli [17]3 years ago
3 0
Here’s the answer!!!!!!

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What is 0.79 (with the repeated 79) in fraction form?
s344n2d4d5 [400]
Expressed a proper fraction in its simplest form, 0.79 is equal to 79/100, or this seventy-nine hundredths
5 0
3 years ago
A ternary digit is either 0, 1, or 2. how many sequences of eight ternary digits containing a single 2 and a single 1 are possib
PIT_PIT [208]

Answer:

56

Step-by-step explanation:

If there can be only one 1 and only one 2, the remaining digits must all be 0. The digits 1 and 2 can be anywhere in the 8 digits, and can be in either order.

There are 8 possible locations in the sequence for the 1, then 7 possible locations for the 2. The total number of possibilities is 8·7 = 56.

12000000, 21000000, 10200000, 20100000, ..., 00000012, 00000021.

8 0
3 years ago
8 less than 4 times the number c is 20 what is the value of c
Brut [27]
Answer is 7


Work

4C-8= 20
+8
4C=28

28/4=7
8 0
3 years ago
The total​ cost, after using a 55cents​-off ​coupon, is ​$3.30. If c represents the cost of the fruit in dollars per​ pound, wha
Rudik [331]
T= total cost= $3.30
c= cost of fruit per pound
x= # of pounds of fruit purchased

EQUATION
T= (c)(x) - $0.55 coupon

to find c, substitute $3.30 total for T
$3.30= cx - $0.55
add $0.55 to both sides
$3.85= cx
divide both sides by x
$3.85/x= c

Hope this helps! :)
5 0
3 years ago
Find all n∈IN such that the number of digits of n equal to n
Butoxors [25]
Note that \log_{10}1=0, \log_{10}10=1, \log_{10}100=2, and so on. The function \log_{10}x is continuous and increasing for all x>0, so when 1, we have 0; when 10, 1; and so on.

This means

\lfloor\log_{10}x\rfloor=\begin{cases}0&\text{for }1\le x

which means we can capture the number of digits of n\in\mathbb N with the function \lfloor\log_{10}n\rfloor+1.

So the problem is the same as finding positive integer solutions to

\lfloor\log_{10}n\rfloor+1=n

We know that n=1 has one digit, so clearly this must be a solution. We need to show that this is the only solution.

Recall that \dfrac{\mathrm d}{\mathrm dx}[\log_{10}x+1]=\dfrac1{\ln10\,x}, while \dfrac{\mathrm d}{\mathrm dx}[x]=1. This means \log_{10}x increases at a much slower rate than x as x\to\infty. We know the two functions intersect when x=1. Therefore it's clear that x>\log_{10}x+1 for all x>1.

Now, it's always the case that \lfloor f(x)\rfloor\le f(x), so we're essentially done:

x>\log_{10}x+1\ge\lfloor\log_{10}x\rfloor+1

which means there are no other solutions than n=1.
5 0
3 years ago
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