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3 years ago

How is naming a line segment different from naming a line?

1 answer:

Studentka2010 [4]3 years ago

7 0

A line is infinite, it has no beginning or end. it's length is infinite.
It can only be named by the equation or a variation of y=mx+c where m is the gradient and c is the y intercept

A line segment has a specific start and end, and a length.
It can be expressed giving two sets of coordinates, (the start and finish)

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URGENT!!!!!! URGENT!!!! WILL MARK AS BRAINLIEST!!!! Benjamin considers purchasing a car for $35,000. He has been approved for a

hram777 [196]

Answer: $1250

Step-by-step explanation:

Given data:

Cost of car = $35,000

Down payment made = $2500

Interest rate = 5.25%

Duration of loan = 6 years

Loan Amount = $3000

Solution:

I = prt/100

where

I = interest

p = principal ($30,000)

r = 5.25%

t = 6

I = $30,000 * 5.25 * 6 /100

= $9450

interest payable on the loan in 6 years is $9450.

Rental

=$600 monthly * 6(12)

= $43,200

Amount made from renting the car is $43,200.

• Money saved by Benjamin in 6 years

= $43,200 – $32500 + $9450

= $43,200 – $41,950

= $1250

Benjamin would have saved $1250 in 6 years.

7 0

2 years ago

Hallamos precios justos empleando funciones cuadráticas (día 3)

lidiya [134]

Answer:

1. See below

2. F(x) = -0.10x² + 8x + 200

3. You earn the maximum income of S/320

4. The income decreases

5. x < 0; x > 100

Step-by-step explanation:

1. Strategies

Let x = the days

Then 100 - x = the kilograms of oranges each day

and 0.10x = the daily price increase

and 2 + 0.10x = the daily price

Income = (kilograms of oranges) × (price per kilogram)

F(x)= (100 - x)(2 + 0.10x)

2. Another expression for F(x)

You could multiply the factors to get a polynomial in its standard form.

$\begin{array}{rcl}F(x) & = & (100 - x)(2 + 0.10x)\\& = & 200 + 8x - 0.10x^{2}\\& = &\mathbf{ -0.10x^{2} + 8x + 200}\\\end{array}$

3. If sales are made in 20 da

$\begin{array}{rcl}F(x) & = & (100 - x)(2 + 0.10x)\\& = & (100 - 20)(2 + 0.10\times20)\\& = & 80(2 + 2.0)\\& = &80 \times 4\\&=&\textbf{S/320}\\\end{array}\\\text{Sales made on Day 20 will earn the maximum income of } \textbf{S/320}$

4. If sales are made after 40 da

(a) If sales are made on Day 40

$\begin{array}{rcl}F(x) & = & (100 - x)(2 + 0.10x)\\F(x) & = & (100 - 40)(2 + 0.10\times40)\\& = & 60(2 + 4.0)\\& = &60 \times 6\\&=&\textbf{S/360}\\\end{array}\\$

(b) If sales are made on Day 41

$\begin{array}{rcl}F(x) & = & (100 - 41)(2 + 0.10\times41)\\& = & 59(2 + 4.1)\\& = &59 \times 6.1\\&=&\textbf{S/359.90}\\\end{array}\\\text{The \textbf{income decreases} after Day 40.}$