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Mariana [72]
3 years ago
11

Can someone please answer this multiple choice question it was only 2 I need it right now I have 10 or 20 minutes left please an

swer it correctly please and please please show work

Mathematics
1 answer:
liq [111]3 years ago
4 0

Answer:

Thanks for reporting me

Step-by-step explanation:

I was just trying to help but ok..... :)

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For f(x) =
vovangra [49]

Answer:

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I got most of the answers so contact me fast. I also need ur help with the assignments, but can give the quiz ones in exchange.

4 0
3 years ago
Y=x^2-12x+45 vertex form and coordinate vertex
zysi [14]
Best Answer

<span><span> x2-12x-45=0</span> </span>Two solutions were found :<span> x = 15 x = -3</span>

Step by step solution :<span>Step  1  :</span>Skip Ad
Trying to factor by splitting the middle term

<span> 1.1 </span>    Factoring <span> x2-12x-45</span> 

The first term is, <span> <span>x2</span> </span> its coefficient is <span> 1 </span>.
The middle term is, <span> -12x </span> its coefficient is <span> -12 </span>.
The last term, "the constant", is <span> -45 </span>

Step-1 : Multiply the coefficient of the first term by the constant <span> <span> 1</span> • -45 = -45</span> 

Step-2 : Find two factors of  -45  whose sum equals the coefficient of the middle term, which is  <span> -12 </span>.

<span><span>     -45   +   1   =   -44</span><span>     -15   +   3   =   -12   That's it</span></span>


Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -15  and  3 
                     <span>x2 - 15x</span> + 3x - 45

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x • (x-15)
              Add up the last 2 terms, pulling out common factors :
                    3 • (x-15)
Step-5 : Add up the four terms of step 4 :
                    (x+3)  •  (x-15)
             Which is the desired factorization

<span>Equation at the end of step  1  :</span> (x + 3) • (x - 15) = 0 <span>Step  2  :</span>Theory - Roots of a product :

<span> 2.1 </span>   A product of several terms equals zero.<span> 

 </span>When a product of two or more terms equals zero, then at least one of the terms must be zero.<span> 

 </span>We shall now solve each term = 0 separately<span> 

 </span>In other words, we are going to solve as many equations as there are terms in the product<span> 

 </span>Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

<span> 2.2 </span>     Solve  :    x+3 = 0<span> 

 </span>Subtract  3  from both sides of the equation :<span> 
 </span>                     x = -3 

Solving a Single Variable Equation :

<span> 2.3 </span>     Solve  :    x-15 = 0<span> 

 </span>Add  15  to both sides of the equation :<span> 
 </span>                     x = 15 

Supplement : Solving Quadratic Equation Directly<span>Solving <span> x2-12x-45</span>  = 0 directly </span>

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

<span> 3.1 </span>     Find the Vertex of   <span>y = x2-12x-45

</span>Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero).<span> 

 </span>Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.<span> 

 </span>Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.<span> 

 </span>For any parabola,<span>Ax2+Bx+C,</span>the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   6.0000 <span> 

 </span>Plugging into the parabola formula   6.0000  for  x  we can calculate the  y -coordinate :<span> 
 </span><span> y = 1.0 * 6.00 * 6.00 - 12.0 * 6.00 - 45.0 
</span>or   y = -81.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : <span> y = x2-12x-45</span>
Axis of Symmetry (dashed)  {x}={ 6.00} 
Vertex at  {x,y} = { 6.00,-81.00}  
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-3.00, 0.00} 
Root 2 at<span>  {x,y} = {15.00, 0.00}</span>

3 0
3 years ago
What is the equation of the line that passes through the point (-4,-2) and has a<br> slope of -1/2?
Soloha48 [4]

Answer:

y =  -  \frac{1}{2} x

8 0
3 years ago
Help with this question asap
Softa [21]

Answer:

ruiwghyrvefufcfuerfyurtgreufru

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Consider the differential equation:
Wewaii [24]

(a) Take the Laplace transform of both sides:

2y''(t)+ty'(t)-2y(t)=14

\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s

where the transform of ty'(t) comes from

L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)

This yields the linear ODE,

-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s

Divides both sides by -s:

Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}

Find the integrating factor:

\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C

Multiply both sides of the ODE by e^{3\ln|s|-s^2}=s^3e^{-s^2}:

s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}

The left side condenses into the derivative of a product:

\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}

Integrate both sides and solve for Y(s):

s^3e^{-s^2}Y(s)=7e^{-s^2}+C

Y(s)=\dfrac{7+Ce^{s^2}}{s^3}

(b) Taking the inverse transform of both sides gives

y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that \frac{7t^2}2 is one solution to the original ODE.

y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7

Substitute these into the ODE to see everything checks out:

2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14

5 0
3 years ago
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