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4 years ago

P=2 (w+h) h the subject

Mathematics

2 answers:

Mumz [18]4 years ago

8 0

P=2(w+h)
P=2w+2h
P-2w=2h
(P-2w)/2 = h
or h=(P-2w)/2

UkoKoshka [18]4 years ago

7 0

P=2(w+h)
remember you can do anythig to an equaotn as lng as you do it to both sides
divide both sides by 2
P/2=w+h
subtract w
(P/2)-w=h

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The function f(x) is a quadratic function and a limited table of values is provided below. Write the equation of the quadratic p

OverLord2011 [107]

Answer:

f(x) = 3x^2 -21x +36

Step-by-step explanation:

The table gives the x- and y-intercepts, which are sufficient to write the equation in factored form. The x-intercepts of 3 and 4 tell you that factors are (x -3)(x -4). When x=0, this product is (-3)(-4) = 12, but the y-intercept value is 3 times that: 36. So, the factored equation is ...

f(x) = 3(x -3)(x -4)

Multiplying this out, we get ...

f(x) = 3(x^2 -7x +12)

f(x) = 3x^2 -21x +36

4 0

3 years ago

A. f(x) = l-2xl - 3 Domain: {1, 2, 3} Find the range

gizmo_the_mogwai [7]

Hello there!

We are given the function:

$\displaystyle \large{ f(x) = | - 2x| - 3} \\$

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

x = 1

$\displaystyle \large{ f(1) = | - 2(1)| - 3} \\ \displaystyle \large{ f(1) = | - 2| - 3} \\$

Recall that any numbers in absolute value are always positive.

$\displaystyle \large{ f(1) = 2 - 3} \\ \displaystyle \large{ f(1) = - 1} \\$

x = 2

$\displaystyle \large{ f(2) = | - 2(2)| - 3} \\ \displaystyle \large{ f(2) = | - 4 | - 3} \\ \displaystyle \large{ f(2) = 4 - 3} \\ \displaystyle \large{ f(2) = 1}$

x = 3

$\displaystyle \large{ f(3) = | - 2(3)| - 3} \\ \displaystyle \large{ f(3) = | - 6| - 3} \\ \displaystyle \large{ f(3) = 6 - 3} \\ \displaystyle \large{ f(3) = 3}$

Therefore, Range: {-1,1,3}

Let me know if you have any questions!

Topic: Absolute Value Function / Modulus Function

6 0

3 years ago

(y-5)/2=(y+6)/3 solve

MA_775_DIABLO [31]

The solution of the given equation represented in the task content can be determined by first evaluating the LCM and hence is; y= 27.

<h3>What is the solution of the equation as given in the task?</h3>

The given equation is; (y-5)/2=(y+6)/3.

Hence, by multiplying the equation by 6, we have;

3(y-5) = 2(y+6)

3y -15 = 2y +12

y = 27.

Hence, it follows that the value of y which is a solution to the equation is; y = 27.