All categories

3 years ago

The function below is written in vertex form or intercept form. Rewrite them in standard form and show your work.

Mathematics

2 answers:

lora16 [44]3 years ago

8 0

Answer:

The standard form as $y=5x^2+30x+41$

Step-by-step explanation:

Given: A function which is written in vertex form or intercept form.

We have to re-write it in standard form that in terms of

Given $y = 5(x+3)^2-4$

Squaring using $(a+b)^2=a^2+b^2+2ab$ , we get,

$y=5(x^2+9+6x)-4$

Multiply 5 inside , we get,

$y=5x^2+45+30x-4$

Solving further , we get,

$y=5x^2+30x+41$

Thus , we have obtained the standard form as $y=5x^2+30x+41$

olga55 [171]3 years ago

5 0

Answer: The standard form of equation will be

$f(x)=5x^2+30x+41$

Step-by-step explanation:

Since we have given that

The vertex form of equation is given by

$y=5(x+3)^2-4$

We need to find the standard form :

Standard form is written as :

$f(x)=ax^2+bx+c$

So, our equation becomes,

$y =5(x+3)^2-4\\\\y=5(x^2+9+6x)-4\\\\y=5x^2+45+30x-4\\\\y=5x^2+30x+41$

Hence, the standard form of equation will be

$f(x)=5x^2+30x+41$

You might be interested in

The width of a rectangular field is 20 feet less than its length. The area of the field is 12,000 ft? What is the length of the

Scilla [17]

Answer:

The answer is 120 feet.

Step-by-step explanation:

The area of the field (A) is:

A = w · l (w - width, l - length)

It is known:

A = 12,000 ft²

l = w - 20

So, let's replace this in the formula for the area of the field:

12,000 = w · (w - 20)

12,000 = w² - 20

⇒ w² - 20w - 12,000 = 0

This is quadratic equation. Based on the quadratic formula:

ax² + bx + c = 0 ⇒

In the equation w² - 20w - 12,000 = 0, a = 1, b = -20, c = -12000

Thus:

So, width w can be either

Since, the width cannot be a negative number, the width of the field is 120 feet.

6 0

3 years ago

Read 2 more answers

How do I solve for x?

notka56 [123]

Solving:
$\frac{3+4x}{2} + \frac{5-x}{3} = \frac{29}{6}$
Make the Least Common Multiple (2,3,6)
$2,3,6\:|2$
$1,3,3\:|3$
$1,1,1\:|\underline{2*3=6}$

Replace denominators and resolve:
 $\frac{3+4x}{2} + \frac{5-x}{3} = \frac{29}{6}$
$\frac{3(3+4x)}{6} + \frac{2(5-x)}{6} = \frac{29}{6}$
Cancel the dominators
$\frac{3(3+4x)}{\diagup\!\!\!\!6} + \frac{2(5-x)}{\diagup\!\!\!\!6} = \frac{29}{\diagup\!\!\!\!6}$
$3(3+4x) + 2(5-x) = 29$
$9 + 12x + 10 - 2x = 29$
$12x - 2x = 29 - 9 - 10$
$10x = 20 - 10$
$10x = 10$
$x = \frac{10}{10}$
$\boxed{\boxed{x = 1}}\end{array}}\qquad\quad\checkmark$

7 0

4 years ago

HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH

8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

$4+\sqrt{6x}$

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

$4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}$

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

$\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}$