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zzz [600]
4 years ago
8

Solve for X: 3-2(x-1)=x-10

Mathematics
2 answers:
lyudmila [28]4 years ago
6 0
3-2(x-1)=x-10 \\\\ 3-2x+2=x-10 \\\\ -2x+5=x-10 \\\\ -2x-x=-10-5 \\\\ -3x=-15 \\\\ \boxed{x=\frac{-15}{-3}=5}
Rus_ich [418]4 years ago
3 0
3-2(x-1)=x-10\\
3-2x+2=x-10\\
-3x=-15\\
x=5
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A group of 18 people ordered soup and sandwiches for lunch. Each person in the group either ordered one soup or one sandwich . T
kkurt [141]
Let's say a = number of ordered soups, b = number of ordered sandwiches.

Then 4.5a + 7.75b = 113.50, and a and b are integers between 0 and 18 inclusive.

How do we tackle this? If all ordered soup, the cost would be $81, so we'll have at least 4 sandwiches. If all ordered sandwich, the cost would be $139.5, so at most 15 sandwiches were ordered. You also know an even number of sandwiches was ordered, to let the price end at 50 cent.

If you brute-force from 4,6,8,10,12 to 14 sandwiches, you find the answer at 10 sandwiches and 8 soups.

10*7.75 + 8*4.50 = 113.50


8 0
4 years ago
How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
ad-work [718]
Note that powers of 2 can be written in binary as

2^0=1_2
2^1=10_2
2^2=100_2

and so on. Observe that n+1 digits are required to represent the n-th power of 2 in binary.

Also observe that

\log_2(2^n)=n\log_22=n

so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,

2_{10}=10_2\iff\log_2(2^1)=\log_22=1
3_{10}=11_2\iff\log_23=1+(\text{some number between 0 and 1})
4_{10}=100_2\iff\log_24=2

That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of \log_23. We only need the integer part, \lfloor\log_23\rfloor, then we add 1.

Now, 2^9=512, and 999 falls between these consecutive powers of 2. That means

\log_2999=9+\text{(some number between 0 and 1})

which means 999 requires \lfloor\log_2999\rfloor+1=9+1=10 binary digits.

Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write 1_2? Or do you pad this number with 0s to get 10 digits, i.e. 0000000001_2? In the latter case, the answer is obvious; 1000\times10=10^4 total binary digits are needed.

In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
8 0
4 years ago
Which radical expression are equivalent to 4^3/5?
photoshop1234 [79]

Answer:

E. 5√4^3

Step-by-step explanation:


7 0
4 years ago
For which fraction pairs would it be useful to use the benchmarks 0, One-half, and 1 to compare? Check all that apply.
koban [17]

Answer:

,

Step-by-step explanation:

thank my answer

pls mark brilinias

8 0
3 years ago
MARKING BRAINLEIST IF CORRECT AND QUICK
egoroff_w [7]

Answer:

Step-by-step explanation:

D)

Hope that helps!

5 0
3 years ago
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