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3 years ago

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Rewrite the function by completing the square f(x)=x^2-14x+63

2 answers:

Veseljchak [2.6K]3 years ago

8 0

Answer:

f(x) = (x - 7)² - 14

General Formulas and Concepts:

<u>Pre-Algebra</u>

Equality Properties

<u>Algebra I</u>

Standard Form: ax² + bx + c = 0
Vertex Form: f(x) = a(bx - c)² + d
Completing the Square: (b/2)²

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = x² - 14x + 63

<u>Step 2: Rewrite</u>

Separate: f(x) = (x² - 14x) + 63
Complete the Square: f(x) = (x² - 14x + 49) + 63 - 49
Simplify: f(x) = (x - 7)² - 14

anastassius [24]3 years ago

6 0

Answer:

-7 and 14 on khan

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Write an equation of the line shown. Then use the equation to find the value of x when y = 150

Ymorist [56]

Answer:

$y = 10x + 30$

x = 12

Step-by-step explanation:

The slope-intercept form formula, $y = mx + b$ , can be used to write an equation for the line.

Where,

m = slope = $\frac{y_2 - y_1}{x_2 - x_1}$

b = y-intercept, which is the point at which the line intercepts the y-axis. At this point, x = 0.

Let's find the slope (m) using the coordinates of the two points given, (0, 30), (3, 60).

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Let,

$(0, 30) = (x_1, y_1)$

$(3, 60) = (x_2, y_2)$

$m = \frac{60 - 30}{3 - 0}$

$m = \frac{30}{3}$

$m = 10$

y-intercept of the line, b = 30

Equation for the line would be:

$y = 10x + 30$

Using the equation, find x when y = 150.

Simply substitute the value for y in the equation to find x.

$150 = 10x + 30$

Subtract 30 from both sides

$150 - 30 = 10x + 30 - 30$

$120 = 10x$

Divide both sides by 10

$\frac{120}{10} = \frac{10x}{10}$

$12 = x$

x = 12

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3 years ago

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Ratling [72]

Answer:

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Step-by-step explanation:

The Food Pyramid

4 0

3 years ago

F(x)=−7x^2 −x and g(x)=9x^2 −4x , find (f−g)(x) and (f−g)(1)

givi [52]

Answer:

f(x) = -7x² - x

g(x) = 9x² - 4x

To find ( f - g)(x) subtract g(x) from f(x)

That's

( f - g)(x) = - 7x² - x - ( 9x² - 4x)

= -7x² - x - 9x² + 4x

Group like terms

( f - g)(x) = - 7x² - 9x² - x + 4x

<h3>( f - g)(x) = - 16x² + 3x</h3>

To find (f - g)(1) substitute 1 into ( f - g)(x)

That's

(f - g)(1) = - 16(1) + 3(1)

= -16 + 3

<h3>= - 13</h3>

Hope this helps you

6 0

3 years ago

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gulaghasi [49]

Answer:

intresting

Step-by-step explanation:

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3 years ago

Find all pairs of real numbers (a,b) such that (x-a)^2+(2x-b)^2=(x-3)^2+(2x)^2

Mrrafil [7]

Answer:

$(3,0)$

$(\frac{-9}{5},\frac{12}{5})$

Step-by-step explanation:

Let's expand both sides.

I'm going to use the following identity to expand the binomial squared expressions: $(u+v)^2=u^2+2uv+v^2$ or $(u-v)^2=u^2-2uv+v^2$ .

Left-hand side:

$(x-a)^2+(2x-b)^2$

$(x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)$

$x^2-2ax+a^2+4x^2-4bx+b^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)$

$5x^2+(-2a-4b)x+(a^2+b^2)$

Right-hand side:

$(x-3)^2+(2x)^2$

$x^2-2(3)x+9+4x^2$

$x^2-6x+9+4x^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-6x)+9$

$5x^2-6x+9$

Now let's compare both sides.

If we want both sides to appear exactly the same we need to choose values $a$ and $b$ such the following are true equations:

$-2a-4b=-6$

$a^2+b^2=9$

So if we solve the system we can find the values $a$ and $b$ such that the left=right.

Let's solve the first equation for $a$ in terms of $b$ .

Add $2a$ on both sides:

$-4b=-6+2a$

Divide both sides by -4:

$b=\frac{-6+2a}{-4}$

Reduce (divide top and bottom by -2):

$b=\frac{3-a}{2}$

Now let's plug this into second equation:

$a^2+b^2=9$

$a^2+(\frac{3-a}{2})^2=9$

$a^2+\frac{9-6a+a^2}{4}=9$ (I used the identity $(u-v)^2=u^2-2uv+v^2$ )

Multiply both sides by 4 to clear the fractions from the problem:

$4a^2+(9-6a+a^2)=36$

Combine like terms on left hand side:

$4a^2+a^2-6a+9=36$

$5a^2-6a+9=36$

Subtract 36 on both sides:

$5a^2-6a-27=0$

Now let's try to factor.

We are going to try to find two numbers that multiply to be 5(-27) and add to be -6.

5(-27)=(5*3)(-9)=15(-9)=-15(9) while -15+9=-6.

So let's replace $-6a$ with $-15a+9a$ and factor by grouping.

$5a^2-15a+9a-27=0$

$5a(a-3)+9(a-3)=0$

$(a-3)(5a+9)=0$

This implies $a-3=0$ or $5a+9=0$ .

Solving the first is easy. Just ad 3 on both sides to get: $a=3$ .

The second requires two steps. Subtract 9 and then divide by 5 on both sides.

$5a=-9$

$a=\frac{-9}{5}$ .

So let's go back to finding $b$ now that we know the $a$ values.

If $a=3$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-3}{2}=0$ .

So one ordered pair $(a,b)$ that satisfies the equation is:

$(3,0)$ .

If $a=\frac{-9}{5}$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-\frac{-9}{5}}{2}$ .

Let's multiply top and bottom by 5 to clear the mini-fraction.

$b=\frac{15-(-9)}{10}$

$b=\frac{24}{10}$

$b=\frac{12}{5}$

So one ordered pair $(a,b)$ that satisfies the equation is:

$(\frac{-9}{5},\frac{12}{5})$ .

5 0

4 years ago