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a certain computer can perform

10

5

calculations per second. how many calculations can it perform in 10 seconds?

x^2 +23x +49

Step-by-step explanation:

First we find the area of the rectangle as though the small square were not cut out of it

A = (x+10) (2x+5)

Foil

2x^2 +5x+20x+50

2x^2 +25x+50

Then we find the area of the small square

A = (x+1) (x+1)

FOIL

x^2 +x+x+1

x^2 +2x+1

Then we subtract the small square from the large rectangle to find the area of the shaded region

2x^2 +25x+50 - (x^2 +2x+1)

Distribute the minus sign

2x^2 +25x+50 - x^2 -2x-1

x^2 +23x +49

[tex]10^6[/tex] calculations in [tex]10[/tex] seconds (or [tex]1,000,000[/tex] calculations in [tex]10[/tex] seconds)

Step-by-step explanation:

The complete exercise is: "A certain computer can perform 10^5 calculations per second. How many calculations can it perform in 10 seconds".

You need to analize the information provided in the exercise in order to solve it.

Let be "x" the number of calculation a certain computer can perform is ten seconds.

According to the exercise this computer can perform [tex]10^5[/tex] calculations in [tex]1[/tex] second. Then, in order to calculate the amount of calculattion it performs in [tex]10[/tex] seconds, you can set up the following proportion:

[tex]\frac{10^5}{1}=\frac{x}{10}[/tex]

Now, you need to find the value of "x".

To solve for "x" you can multiply both sides of the equation by [tex]10[/tex].

Then, you get:

[tex](10)(\frac{10^5}{1})=(\frac{x}{10})(10)\\\\10^6=x\\\\x=10^6[/tex]

Therefore, the computer can perform [tex]10^6[/tex] calculations in [tex]10[/tex] seconds (or [tex]1,000,000[/tex] calculations in [tex]10[/tex] seconds)

The computer can perform 1051050 calculations in 10 seconds.

Step-by-step explanation:

1. Put the quantity of calculations the computer can perform in 1 second:

[tex]105105\frac{calculations}{1second}[/tex]

2. Multiply the quantity of calculations by 10 seconds:

[tex]105105\frac{calculations}{1second}*10seconds=1051050calculations[/tex]

Note that the units of seconds are cancelled as they appear on the numerator and the denominator, and the final response unit is given in calculations number.

The computer can perform 1051050 calculations in 10 seconds.

Steven is 15 and one half years old now

Step-by-step explanation:

Linear equations

Sometimes we need to know the value of a variable in a given equation and that variable is given as a polynomial of degree 1. Solving for that variable means isolating it and replacing the other known values

Steven was 7 years old eight and one-half years ago. If we call x as the actual age of Steven, his age eight and one-half years ago was

[tex]x-8\frac{1}{2}[/tex]

We know he was 7 years old then, so

[tex]x-8\frac{1}{2}=7[/tex]

We need to know his actual age, so let's solve for x, adding [tex]8\frac{1}{2}[/tex] to each side of the equation

[tex]x-8\frac{1}{2}+8\frac{1}{2}=7+8\frac{1}{2}[/tex]

Simplifying

[tex]x=15\frac{1}{2}[/tex]

Steven is 15 and one half years old now

∠B = 41°

Step-by-step explanation:

∠BCD is an exterior angle of the triangle

The exterior angle is equal to the sum of the 2 opposite interior angles, thus

∠A + ∠B = 113

72 + ∠B = 113 ( subtract 72 from both sides )

∠B = 41°

Start by dividing the numerator and denominator by 2.

12/32 = 6/16

Now that can also be divided by 2.

6/16 = 3/8

That's as far as we can go because 3/8 can not be divided any more.

We could have divided 12/32 by 4 and saved a step.