I,m trying to solve this algebra question:
x^+9x+20=0

Answers

Answer 1

Answer: ${ x }^( 2 )+9x+20=0\n \n \left( x+5 \right) \left( x+4 \right) =0\n \n \therefore \quad x=-5,\quad x=-4.$

Answer 2

Answer: Ok, let assume it's $x^2$

$x^2+9x+20=0\nx^2+4x+5x+20=0\nx(x+4)+5(x+4)=0\n(x+5)(x+4)=0\nx=-5 \vee x=-4$

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The graph of which function has an axis of symmetry at x =-1/4 ?f(x) = 2x2 + x – 1

f(x) = 2x2 – x + 1

f(x) = x2 + 2x – 1

f(x) = x2 – 2x + 1

Answers

we know that

The equation of the vertical parabola in vertex form is equal to

$y=a(x-h)^(2)+k$

where

(h,k) is the vertex

The axis of symmetry is equal to the x-coordinate of the vertex

$x=h$ ------> axis of symmetry of a vertical parabola

we will determine in each case the axis of symmetry to determine the solution

case A) $f(x)=2x^(2)+x-1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+1=2x^(2)+x$

Factor the leading coefficient

$f(x)+1=2(x^(2)+0.5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+1+0.125=2(x^(2)+0.5x+0.0625)$

$f(x)+1.125=2(x^(2)+0.5x+0.0625)$

Rewrite as perfect squares

$f(x)+1.125=2(x+0.25)^(2)$

$f(x)=2(x+0.25)^(2)-1.125$

the vertex is the point $(-0.25,-1.125)$

the axis of symmetry is

$x=-0.25=-(1)/(4)$

therefore

the function $f(x)=2x^(2)+x-1$ has an axis of symmetry at $x=-(1)/(4)$

case B) $f(x)=2x^(2)-x+1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-1=2x^(2)-x$

Factor the leading coefficient

$f(x)-1=2(x^(2)-0.5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-1+0.125=2(x^(2)-0.5x+0.0625)$

$f(x)-0.875=2(x^(2)-0.5x+0.0625)$

Rewrite as perfect squares

$f(x)-0.875=2(x-0.25)^(2)$

$f(x)=2(x-0.25)^(2)+0.875$

the vertex is the point $(0.25,0.875)$

the axis of symmetry is

$x=0.25=(1)/(4)$

therefore

the function $f(x)=2x^(2)-x+1$ does not have a symmetry axis in $x=-(1)/(4)$

case C) $f(x)=x^(2)+2x-1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+1=x^(2)+2x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+1+1=x^(2)+2x+1$

$f(x)+2=x^(2)+2x+1$

Rewrite as perfect squares

$f(x)+2=(x+1)^(2)$

$f(x)=(x+1)^(2)-2$

the vertex is the point $(-1,-2)$

the axis of symmetry is

$x=-1$

therefore

the function $f(x)=x^(2)+2x-1$ does not have a symmetry axis in $x=-(1)/(4)$

case D) $f(x)=x^(2)-2x+1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-1=x^(2)-2x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-1+1=x^(2)-2x+1$

$f(x)=x^(2)-2x+1$

Rewrite as perfect squares

$f(x)=(x-1)^(2)$

the vertex is the point $(1,0)$

the axis of symmetry is

$x=1$

therefore

the function $f(x)=x^(2)-2x+1$ does not have a symmetry axis in $x=-(1)/(4)$

the answer is

$f(x)=2x^(2)+x-1$

axis of symmetry is the x value of the vertex

for
y=ax^2+bx+c
x value of vertex=-b/2a

first one
-1/2(2)=-1/4
wow, that is right

answer is first one
f(x)=2x^2+x-1

A)
be
85°
120°
(2x - 10°
X

Answers

Let's see what to do buddy...

_________________________________

Step (1)

We know that the sum of the interior angles of any n-sided figure is obtained from the following equation :

$(n - 2) * 180$

Our figure has 5 sides ;

So , sume the interior angles of it equals :

$(5 - 2) * 180 = 3 * 180 = 540 \n$

_________________________________

Step (2)

First look at the photo which I post.

According it we have :

$(180 - x) + 120 + 85 + (2x - 10) + 90 = 540 \n$

$2x - x + 475 - 10 = 540$

$x + 465 = 540$

Subtract the sides of the equation minus 465 :

$x = 540 - 465$

$x = 75$

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

Suppose a parabola has vertex (–4, 7) and also passes through the point (–3, 8). Write the equation of the parabola in vertex form.

Answers

The equation for a parabola can also be written in the "vertex form":
$y = a (x-h) ^ 2 + k$
Where,
the vertex of the parabola is the point (h, k).
The value of a is the term that accompanies x ^ 2
Substituting values we have:
$y = a (x - (- 4)) ^ 2 + 7$
Rewriting we have:
$y = a (x + 4) ^ 2 + 7$
For the point (-3, 8) we have:
$8 = a (-3 + 4) ^ 2 + 7$
From here, we clear the value of a:
$8 = a (1) ^ 2 + 7 8 = a + 7 a = 8 - 7 a = 1$
Then, the equation is given by:
$y = (x + 4) ^ 2 + 7$
Answer:
The equation of the parabola in vertex form is:
$y = (x + 4) ^ 2 + 7$

y=a(x-h)^2 +k
in the vertex (h, k) given that vertex (-4, 7)
we get y=(x-(-4))^2 +7
y=(x+4)^2 +7
hope it helps

Write the equation of a line perpendicular to y=2x+8 and goes through the point (2, -2).

Answers

Hi,
y+2=-1/2(x-2) ==>y=-x/2-1.
The slope (if this is right) : 2*s=-1==>s=-1/2

Tom Brown decided to purchase a new bike on an installment loan. The bike was $300.00. He agreed to pay $30 a month for 12 months. What is the finance charge in dollars?The finance charge is