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ladessa [460]
3 years ago
15

"If x is not > 0, then x^2 is not > 10" is the: -

Mathematics
1 answer:
Elanso [62]3 years ago
7 0
<span>If </span><span /><span><span><span><span><span><span>x</span><span>≠</span><span><span><span>0</span></span></span></span><span /></span></span><span /></span><span>x≠0</span></span><span>, then </span><span /><span><span><span><span><span><span><span><span><span><span><span><span><span><span><span><span>x</span><span /></span><span><span>2</span><span /></span></span></span></span><span /></span><span><span><span><span>−</span><span /></span><span><span>−</span><span /></span></span><span /></span><span><span><span>√</span></span><span /></span></span></span><span /></span><span><span>x</span><span /></span><span><span /><span /></span></span></span><span>=</span></span><span /></span></span><span /></span><span>x2x=</span></span>
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What is the slope of the line that passes through the points (-3, 5) and (1, 7)?
just olya [345]

\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{7}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{7-5}{1-(-3)}\implies \cfrac{7-5}{1+3}\implies \cfrac{2}{4}\implies \cfrac{1}{2}

4 0
3 years ago
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Which equation represents a circle that contains the point (-5,-3) and has a center at (-2,1)
MatroZZZ [7]

Answer:

The equation of circle is:

Solution:

The general form of equation is given as:

Where, "r" is radius of circle and (a, b) is center of circle

Given that center at (-2, 1)

a = -2 and b = 1

Substitute (a , b) = (-2, 1) and (x, y) = (-5, -3) in general equation

Substitute r = 5 and (a, b) = (-2, 1) in general equation

Thus the equation of circle is found

4 0
3 years ago
Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

x=t-3

y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

3 0
3 years ago
Read 2 more answers
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