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4 years ago

8

How do I figure this one out whats the answer

1 answer:

Semenov [28]4 years ago

3 0

Look at the picture

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In the year 2000, the average cost of a computer could be modeled by the equation C = -5t2 + 750, where t is the number of years

elena55 [62]

Answer:

$\Delta C=-5t^2-250$

Step-by-step explanation:

Given:

The average cost of a computer in the year 2000 is given as:

$C=-5t^2+750$

The average cost of a computer in the year 2008 is given as:

$C=-10t^2+500$

Now, the difference in average cost between the years 2008 and 2000 can be calculated by subtracting the average cost in 2000 from the average cost in 2008.

Framing in equation form, we get:

Difference in average cost (ΔC) is given as:

$\Delta C=C_{2008}-C_{2000}\\\\\Delta C= (-10t^2+500)-(-5t^2+750)\\\\\textrm{Distributing the megative sign inside the second polynomial, we get:}\\\\\Delta C=-10t^2+500+5t^2-750\\\\\textrm{Grouping like terms, we get}\\\\\Delta C=(-10t^2+5t^2)+(500-750)\\\\\Delta C=-5t^2-250$

Therefore, the difference in the costs for a computer between 2008 and 2000 is $\Delta C=-5t^2-250$

7 0

4 years ago

I need some help

Radda [10]

Answer:

cube

Step-by-step explanation:

prisms are 3d figures just like cubes are

the other options are just 2d figures

4 0

3 years ago

Read 2 more answers

49–2ax –a^2–x^2 factor

vampirchik [111]

49–2ax –a^2–x^2

Extract the negative sign

= 49 - (a^2 +2ax +x^2)

Factor using a^2 + 2ab + b^2 = (a+b)^2

= 49 - (a+x)^2

Factor using a^2 - b^2 = (a-b)(a+b)

= ( 7- (a + x)) * (7 + (a + x))

= ( 7- a - x) * (7 + a + x)

4 0

3 years ago

Which graph represents the translation

Arte-miy333 [17]

The second one. The absolute value sign makes all the y values positive, meaning the negative 4 will turn into positive 4.

3 0

3 years ago

shakira invests $500000 at 5% compound intwerest per annum . calculate how many years it takes for the value to double in value?

Kay [80]

Answer:

$t=14.2\ years$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=?\ years\\ P=\$500,000\\ r=0.05\\n=1\\ A=\$1,000,000$

substitute in the formula above and solve for t

$1,000,000=500,000(1+\frac{0.05}{1})^{(1)t}$

$2=(1.05})^{t}$

Apply log both sides

$log(2)=log[(1.05})^{t}]$

$log(2)=(t)log(1.05)$

$t=log(2)/log(1.05)$

$t=14.2\ years$

8 0

3 years ago