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malfutka [58]
3 years ago
12

What is a counterexample to this claim?

Mathematics
1 answer:
yan [13]3 years ago
5 0
Answer: B
Explanation: 9 is the only odd number meaning it’s contradicting the claim that all perfect squares are even since 9 isn’t even. a 3x3 square would be 9 but its also perfect explaining how not all perfect squares are even.
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What is 1+2+3+4+5+6+7+8+9+100=????????
Leya [2.2K]

Answer:

145

Step-by-step explanation:

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3 years ago
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Which ordered pair is the solution to the system of equations? {x+3y=-12, y= 1/3x-6?
Julli [10]

For this, I'll be using the substitution method. Since y = 1/3x - 6, replace the y variable in the first equation with 1/3x - 6 and solve for x:

x+3(\frac{1}{3}x-6)=-12\\ x+x-18=-12\\ 2x-18=-12\\ 2x=6\\ x=3

Now that we have the value of x, plug it into either equation to solve for y:

3+3y=-12\\ 3y=-15\\ y=-5\\ \\ y=\frac{1}{3}*3-6\\ y=1-6\\ y=-5

<h3><u>In short, the solution to this equation is (3,-5).</u></h3>
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3 years ago
What are the steps to construct parallel lines using a compass and straightedge.
kari74 [83]

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Let f(x) = 3x+2 and g(x)=6x-7 find f(x)-g(x)
makkiz [27]
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3 0
4 years ago
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Find the equation for the circle with a diameter whose endpoints are (3,-1) and (-2,-1)
cricket20 [7]

Answer:

The equation for the circle is: (x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2

Step-by-step explanation:

Here, let us assume the two end points of the diameter are:

P (3,-1) and Q (-2,-1)

Now, as we knew diameter is the chord that passer through the center of the circle.

⇒ Mid point of DIAMETER  = Center coordinates  of the Circle

Let us assume O(x,y) is the center of the line  segment PQ.

So, by MID POINT FORMULA:

(x,y)= (\frac{3 -2}{2}  , \frac{-1-1}{2} )  =( \frac{1}{2},\frac{-2}{2})  \\\implies (x,y) =( 0.5  , -1)

⇒ The center coordinates of the circle  = O(0.5,-1)   ...... (1)

Now, RADIUS = Half of DIAMETER

Using DISTANCE FORMULA:

PQ  = \sqrt{(3-(-2))^2 + (-1 - (-1))^2}   = \sqrt{(5)^2 + 0}  = 5

So, Radius  = 5/2  = 2.5 units

Now, the  equation of circle with radius r and center coordinate ( h,k) is given as: (x-h)^2 + (y-k)^2 = r^2

Substitute r = 2.5 and (h,k)  = (0.5,0) we get:

(x-0.5)^2 + (y-0)^2 = (2.5)^2\\\implies (x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2

Hence the equation for the circle is: (x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2

5 0
4 years ago
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