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KatRina [158]
2 years ago
7

Enter an equation in point-slope form

Mathematics
1 answer:
Burka [1]2 years ago
8 0

Answer:

y = 2/5x + 39/5

Step-by-step explanation:

point slope form is y = mx + b

so far we have y = 2/5x + b

m is slope which is given. We need to find b which is y-intercept.

They give us the point (-2,7). We can plug this into the equation to find b.

7 = 2/5(-2) + b

7 = -4/5 + b

b = 39/5

The equation now looks like y = 2/5x + 39/5

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Anothony has 24 squash plants and 18 corn seedlings that he plans to plant in rows on his vegetable garden. Each row will contai
sleet_krkn [62]

Anthony needs to find the greatest common factor of both 18 and 24. The reason why is because this way he can have the same number of plants in each row while having the greatest number in each row possible. The amount of plants he needs to put in each row is 6.

4 0
3 years ago
A 16oz. can of Creamy Corn sells for $1.09, while the 32oz. can is on sale for $1.89. How much would you save with the sale pric
Reptile [31]
80 cents Because $1.89 minus $1.09 equals 80 cents
4 0
3 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
3 years ago
X over 3 is less then or equal to 5
Trava [24]

Answer:

x/3 ≤ 5

I'm not sure if this is what you were looking for. If not, please be more specific in your questions!

7 0
2 years ago
Read 2 more answers
An ice cream shop uses the following ingredients to make 1 sundae.
Sedbober [7]

Answer:

8 sundaes

4:32

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3 0
2 years ago
Read 2 more answers
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