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NikAS [45]
3 years ago
10

How to turn negative into mixed numbers into improper fractions

Mathematics
1 answer:
masya89 [10]3 years ago
4 0
U turn negative numbers from mixed numbers to improper fractions the same way u do with positive...u just temporarily disregard the negative sign.

Example : 
- 2 2/3.....temporarily disregard the negative sign......take ur whole number (2) and multiply it by ur denominator (3) giving u 6, then add that to ur numerator (2) giving u 8....then put that number over the original denominator (3) giving u 8/3...and then put back ur negative sign...-8/3
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If you increase your bicycle speed by 5.6 miles per hour, the speed will be 17.9 miles per hour. What is your bicycle speed now?
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Answer:

12.3 mph

Step-by-step explanation:

17.9 - 5.6 = 12.3

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Problems<br>Simplity each expression 4(9+8)​
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36+32 << multiply

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4{[5 (x-3)+2]-3 [2 (x+5)-9]}
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parenthasees exponents mult/division additon/subtraction

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x-3 and x+5 we can't do anything with so next
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(2x+10-9)=2x+1

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3 years ago
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A circle has the equation 2x²+12x+2y²−16y−150=0.
KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

(x-h)^{2} +(y-k)^{2}=r^{2} (1)

Where (h,k) are the coordinates of the center and r is the radius.

Now, we are given the equation of this circle as follows:

2x^{2}+12x+2y^{2}-16y-150=0 (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

2(x^{2}+6x+y^{2}-8y-75)=0 (3)

Rearranging the equation:

x^{2}+6x+y^{2}-8y=75 (4)

(x^{2}+6x)+(y^{2}-8y)=75 (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of (a\pm b)^{2}=a^{2}\pm+2ab+b^{2}:

<u>For the first parenthesis:</u>

x^{2}+6x+b^{2}

We can rewrite this as:

x^{2}+2(3)x+b^{2}

Hence in this case b=3 and b^{2}=9:

x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}

<u>For the second parenthesis:</u>

y^{2}-8y+b^{2}

We can rewrite this as:

y^{2}-2(4)y+b^{2}

Hence in this case b=-3 and b^{2}=9:

y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}

Then, equation (5) is rewritten as follows:

(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16 (6)

<u>Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.</u>

Rearranging:

(x-3)^{2}+(y-4)^{2}=100 (7)

At this point we have the circle equation in the center radius form (x-h)^{2} +(y-k)^{2}=r^{2}

Hence:

h=-3

k=4

r=\sqrt{100}=10

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