$\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] ~\dotfill\\\\ g(x)=3(x-\stackrel{h}{2})^2+(\stackrel{k}{-5})\qquad \qquad \stackrel{\textit{vertex}}{(2,-5)}$

3 0

3 years ago

Evaluate the expression below.

o-na [289]

Answer:

Step-by-step explanation:

Use:

|a| = a for a ≥ 0 or |a| = -a for a < 0

PEMDAS:

P Parentheses first

E Exponents

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

========================================

|8 - 7| · 4 + 3(-1)

= |1| · 4 - 3

= 1 · 4 - 3

= 4 - 3

= 1

6 0

2 years ago

Read 2 more answers

Find the sum and express it in the simplest form. (2x^2+3x-4)+(x^2-x-2)

Katen [24]

18 hope it helped!!

:) be my pleasure!

tysm

8 0

2 years ago

If (x, y) is a solution to the system of equations, what is the value of y? 1/4 x + 1/8 y = 2 1/3 x + 1/2 y = 4 A) 4 B) 6 C) -6

andre [41]

<h3>The value of y is 4</h3>

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

$\frac{1}{4}x + \frac{1}{8}y =2 ---------- eqn\ 1\\\\\frac{1}{3}x + \frac{1}{2}y = 4 -------------- eqn\ 2$

We have to find value of y

From eqn 1,

$\frac{1}{4}x + \frac{1}{8}y =2 \\\\2x + y = 2 \times 8\\\\2x + y = 16 ---- eqn\ 3$

From eqn 2,

$\frac{1}{3}x + \frac{1}{2}y = 4\\\\2x + 3y = 4 \times 6\\\\2x + 3y = 24 ------ eqn\ 4$

Subtract eqn 3 from eqn 4

2x + 3y = 24

2x + y = 16

( - ) -----------------

2y = 8

y = 4

Thus value of y is 4

5 0

2 years ago