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2 years ago

5

Solve the equation by using the quadratic formula.

2 answers:

Luda [366]2 years ago

7 0

Answer:

The answer is B

Step-by-step explanation:

Sergeeva-Olga [200]2 years ago

4 0

Answer:

B

Step-by-step explanation:

2x² + 5x -3 = 0

2x² - x + 6x - 3 = 0

x(2x-1) + 3(2x-1) = 0

(x+3) (2x-1) = 0

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Label each triangle

Ne4ueva [31]

Answer: Top right and left are ACUTE, Bottom right is RIGHT, and bottom left is OBTUSE.

4 0

3 years ago

Read 2 more answers

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

Find the inner product for (3,1,4)*(2,8,2) and state whether the vectors are perpendicular.

faltersainse [42]

Good evening ,

Answer:

<h2>•• Inner product = 3×2 + 1×8 + 4×2 = 22</h2><h2 /><h2>•• since 22 ≠ 0 then the vectors are not perpendicular.</h2>

:)

5 0

3 years ago

Read 2 more answers

What is (13-2x)(13-2x) ?

Irina-Kira [14]

(13-2x)(13-2x)

(13+-2x)(13+-2x)

(13)(13)+(13)(-2x)+(-2x)(13)+(-2x)(-2x)

169-26x-26x+4x^2

4x^2-52x+169

I hope that's help !

I

3 0

2 years ago

Read 2 more answers

What is the answer to the question

S_A_V [24]

The equation of the parallel line is y = –2x – 11.

Solution:

Given equation of the line is y = –2x – 5.

Slope of this line is $(m_1)$ = –2

To write the equation parallel to this line and passes through (–4, –3).

<em>If two lines are parallel, then they have the same slope.</em>

$\Rightarrow m_1=m_2$

$\Rightarrow m_2 =-2$

<u>Point-slope formula:</u>

$y-y_1=m(x-x_1)$

Substitute the given values in the formula, we get

$y-(-3)=-2(x-(-4))$

$y+3=-2(x+4)$

$y+3=-2x-8$

Subtract 3 from both sides of the equation.

$y+3-3=-2x-8-3$

$y=-2x-11$

Hence the equation of the parallel line is y = –2x – 11.

5 0

3 years ago