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lisov135 [29]
4 years ago
6

Which equations can be used to solve for y, the length of the room? Check all that apply. y(y + 5) = 750 y2 – 5y = 750 750 – y(y

– 5) = 0 y(y – 5) + 750 = 0 (y + 25)(y – 30) = 0
Mathematics
2 answers:
Alex777 [14]4 years ago
5 0
Let's assume length of the room as y
so the width of the room = y - 5

Area
y * (y - 5) = 750
Elodia [21]4 years ago
4 0

Answer with explanation:

We have to find ,those equation which is used to solve for, y ,the length of room.

Measurement of length is not given,but surely it will be a positive Integer , not equal to 0.

We will check from option 1,which of these equation gives positive Integral value.

1. y×(y+5)=750

→y² + 5 y - 750=0

→y² + 30 y - 25 y - 750=0

→y× (y +30) -25 × (y +30)=0

→ (y - 25)(y + 30)=0

→ y -25 =0  ∧ y + 30=0

y= 25 ∧ y = -30

It gives a Positive value.So, this equation can be used to solve for y, the length of the room.

2. y² - 5 y =750

→y² - 5 y - 750=0

→y² - 30 y + 25 y - 750=0

→y× (y -30) +25 × (y -30)=0

→ (y + 25)(y - 30)=0

→ y +25 =0  ∧ y - 30=0

y= -25 ∧ y = 30

It gives a Positive value.So, this equation can be used to solve for y, the length of the room.

3. 750 - y×(y-5)=0

⇒750 - y² +  5 y=0

⇒-1× (y² - 5 y - 750)=0

→y² - 5 y - 750=0

→y² - 30 y + 25 y - 750=0

→y× (y -30) +25 × (y -30)=0

→ (y + 25)(y - 30)=0

→ y +25 =0  ∧ y - 30=0

y= -25 ∧ y = 30

It gives a Positive value.So, this equation can be used to solve for y, the length of the room.

4. y × (y -5) + 750 =0

y² - 5 y + 750=0

you will not get real root,so this equation can't be used  to solve for y, the length of the room.

5. (y + 25)(y-30)=0

→y+ 25 =0 ∧→ y -30=0

y= -25 ∧→ y =30

It gives a Positive value.So, this equation can be used to solve for y, the length of the room.

→→Equation 1,2,3,and 5 that is apart from equation 4, all of equation can  be used to solve for y, the length of the room.

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How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
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Note that powers of 2 can be written in binary as

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Also observe that

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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
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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.
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What is the simplified form of x minus 2 over x squared plus x minus 6divided byx squared plus 5x plus 4 over x plus 4?
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Answer:

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Step-by-step explanation:

You have asked for the simpilfied form of ...

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That would be ...

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We suspect that's not what you meant. Parentheses are required for grouping when you write math expressions in text form. They work best using math symbols instead of words. We think you mean

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<em>Comment on simplifying rational expressions</em>

Division of fractions works the same whether you're working with numbers or polynomials (or anything else). Dividing by something is the same as multiplying by its inverse (reciprocal).

(a/b)/(c/d) = (a/b)·(d/c)

I learned this as "invert and multiply". I've recently seen it referred to as "copy dot flip", meaning you copy the numerator, use a dot symbol to indicate multipication, then flip the denominator (make its reciprocal) to become what you're multiplying by.

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