What is the value of |-6|—|6|-(-6)?<br><br> The solution is

Algebra 1 help!!!!!!!!!

baherus [9]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of expo functions.

f(x) = -8(2)^x - 12 ,

for f(0) ,here substitute x = 0

so we get as ,

==> f(0) = -8(2)^0 -12

==> f(0) = -8-12

==> f(0) = -20

hence, f(0) = -20

6 0

3 years ago

Read 2 more answers

F(x) = 2x3 + 732 – 4x - 5<br> g(x) = 3x – 2<br> Find ( f - g)(x).

Nadya [2.5K]

Answer:

(f-g) (x)=731.............

6 0

3 years ago

I really need help! This is last minute and its 1 am. I'm tired so I'm going to leave it to you guys to solve my problems and pr

Illusion [34]

Answer:

<h2><u>question 1</u></h2>

$n^{2} - 20n -96 = 0$

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence <u>n = 24 and n = -4 </u>

<u></u>

<h2><u>Question 2 </u></h2>

in the form $ax^{2} +bx +c = 0$

= $x^{2} +12x - 48$

make use of the formula :

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}$

replace values to make 2 equations :

1. $\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}$ = 3.17

2. $\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}$ = -15.2

hence <u>x = 3.17 and x = -15.2</u>

<u />

<h2><u>Question 3 </u></h2>

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence<u> x = 10 and x = 4</u>

<u />

<h2><u>Question 4 </u></h2>

in the form $ax^{2} +bx +c = 0$

this becomes $5b^{2} -20b-18-7$

= $5b^{2} -20b-25$

can simplify by 5

= $b^{2} -4b-5 =0\\$

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

hence <u>b = 5 and b = -1</u>

5 0

3 years ago

Read 2 more answers

Suppose that a company needs 1, 200,000 items during a year and that preparation for each production run costs $500. Suppose als

MAVERICK [17]

Answer:

The number of items in each production run so that the total costs of production and storage are minimized is 8165 items/run

Step-by-step explanation:

We will use the following variables:

Q = Quantity being ordered

Q* = the optimal order Quantity: the result being sought

D = annual Demand for the item, over the year

P = unit Production cost

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run)

H = annual cost to Hold one unit

It is important to note which variables are annualized, which are per-order and which are per-unit.

Using the variables, here are the components of the first equation

Total Cost, TC = PC + SC + HC

PC = P x D : Production Cost = unit Production cost times the annual Demand

SC = (D x S)/Q : Setting up Cost = annual Demand times cost per production setup, divided by the order Quantity (number of units)

HC = (H x Q)/2: Holding Cost = annual unit Holding cost times order Quantity (number of units), divided by 2 (because throughout the year, on average the warehouse is half full).

So TC = PC + SC + HC = (P x D) + ((D x S)/Q) + ((H x Q)/2) = PD + (DS/Q) + HQ/2

To obtain the optimal order quantity, Q* that minimizes TC, at the minimum TC, dTC/dQ = 0

dTC/dQ = (H/2) – (D x S)/(Q²) = 0

(H/2) – (D x S)/(Q²) = 0

Solving for Q, which is Q* at this point.

(Q*)² = 2DS/H

Q* = √(2DS/H)

D = annual demand for the item = 200000

S = cost of setting up a production run, regardless of the number of units in the production run (fixed cost per production run) = $500

H = annual cost to Hold one unit = $3

Q* = √(2×200000×500/3) = 8164.97 = 8165 items.

3 0

3 years ago

V150<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B150%20%7D%20%20%5C%5C%20" id="TexFormula1" title=" \sqrt{150 } \\ " alt=

Anuta_ua [19.1K]

Answer: 12.2474487139

Step-by-step explanation:

(I think what you wrote is just the square root of 150? if I read this wrong please let me know so I can correct my answer.)

*If you plug in the square root of 150 into a calculator, you get 12.2474487139

6 0

3 years ago

What is the value of |-6|—|6|-(-6)? The solution is