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3 years ago

Ahh help!!

Mathematics

1 answer:

Rzqust [24]3 years ago

3 0

Answer:

I think its 678.24

Hope it helps!

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See above to see question

andrew-mc [135]

Answer:

A)65°

B)110°

C)50°

D)55°

E)80°

Step-by-step explanation:

Since the angle on a straight line should be 180, you simply subtract the given angles from 180

hope I helped

8 0

3 years ago

I needa know how to do this

Elena L [17]

Answer:

So I (3,0) becomes I' (1,3)

H (5,2) H' (1,1)

G (2,4) G' (2,1)

Step-by-step explanation:

So let's start with I, just count how many times you move to the left then how many times you move up.

UR WELCOME BABEEEE~ <3

5 0

3 years ago

Which expression is equivalent to (1/16)^-4

Dmitry_Shevchenko [17]

Answer: =65536

Step-by-step explanation: hope this help

4 0

3 years ago

Read 2 more answers

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

What is the answer 2c+(3c-4)=a-4

gavmur [86]

Answer:idk

Step-by-step explanation:

5 0

3 years ago