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3 years ago

Someone please help!!!

Mathematics

1 answer:

ycow [4]3 years ago

8 0

Answer:

C: -1,2

Step-by-step explanation:

The solution is where both line intersect and touch each other.

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4) John has a balance of $1,500 in his First State Bank checking account. He

Anni [7]

Answer:

2036.81

Step-by-step explanation:

1500+477.29+237.52+550=2764.81

2764.81-728=2036.81

4 0

2 years ago

Trent is saving money for a new

Dafna11 [192]

He saves 10$ each week

7 0

3 years ago

Provide a counterexample for the following statement: "The product of two irrational numbers is always irrational."

gulaghasi [49]

We know that root 2 is irrational, but root 2 times root 2 is 2, which is rational.

Hope this helped!!!

6 0

3 years ago

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

A. 4 feet by 1 1/2 feet

Triss [41]

The answer would be C because 2*2 =4 and 7/8*2 =1.75 put that into fraction form. 1 3/4

5 0

3 years ago