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3 years ago

Find the limit as t approaches 8 for the function f(t)=7(t-1)(t-1)

Mathematics

1 answer:

Ratling [72]3 years ago

5 0

Answer:

7³

Step-by-step explanation:

You could answer this after a quick inspection. 7 remains constant as t approaches 8. Each of the (t -1) terms will approach 7. So, multiplying these factors together, you'll get 7(7)(7) = 7³3 as your answer, the limit as t approaches 8 of the given expression.

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9. Determine the equation of a line that passes though the points (3,-4) and (6,2). [

stira [4]

Answer:

y=2x-10

Step-by-step explanation:

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

First, let's find what m is, the slope of the line...

The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (3,-4), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=3 and y1=-4.

Also, let's call the second point you gave, (6,2), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=6 and y2=2.

Now, just plug the numbers into the formula for m above, like this:

2 - -4

6 - 3

or...

m=2

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

3 0

3 years ago

Need help with this question

ELEN [110]

Step-by-step explanation:

coordinate point of B(2,5)

and coordinate point ofC(3,1)

Distance of BC

$\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

$= \sqrt{ {(3 - 2)}^{2} + {(1 - 5)}^{2} }$

$= \sqrt{ {1}^{2} + {( - 4)}^{2} }$

$= \sqrt{1 + 16} = \sqrt{17} = 4.12 = 4.1$

5 0

3 years ago

Evaluate the expression when a<br> -4 and b= 7.<br> a-9b

Shalnov [3]

$\huge\text{Hey there!}\\\\\large\text{a - 9b}\\\large\text{= -4 - 9(7)}\\\large\text{= -4 - 63}\\\large\text{= -67}\\\\\large\boxed{\rm{Therefore, your\ answer: -67}}\huge\checkmark\\\\\\\large\text{Good luck on your assignment and enjoy your day!}}\\\\\frak{Amphitrite1040:)}$

8 0

3 years ago

Diana invested $3000 in a savings account for 3 years. She earned $450 in interest over that time period. What interest rate did

matrenka [14]

If I=P*r*t, we can input 450 in for I, 3000 in for the principal (P, or the starting rate), and 3 for t=time because there were three years, resulting in 450=9000*r. Next, we can divide both sides by 9000 to get 0.05 or 5% as our answer.

8 0

3 years ago

Determine how many different computer passwords are possible if (a) digits and letter so can be repeated and (b) digits and lett

Lerok [7]

Answer:

a) 1,188,137,600 different passwords.

b) 710,424,000 different passwords.

Step-by-step explanation:

We have a code of 2 digits followed by 5 letters.

First, the total number of digits is 10

The total number of letters is 26.

Then:

a) Digits and letters can be repeated.

Here we need to count the number of options for each selection.

For the first digit, we have 10 options.

For the second digit, we have 10 options.

For the first letter, we have 26 options.

For the second letter, we have 26 options.

For the third letter, we have 26 options.

For the fourth letter, we have 26 options.

For the fifth letter, we have 26 options.

The total number of combinations will be equal to the product of the number of options. We get:

Combinations = 10*10*26*26*26*26*26 = (10^2)*(26^5) = 1,188,137,600

This means that we have 1,188,137,600 different possible passwords.

b) Digits and letters can not be repeated.

We start in the same way as above:

For the first digit, we have 10 options.

For the second digit, we have 9 options, because one is already used.

For the first letter, we have 26 options.

For the second letter, we have 25 options, because one letter is already used.

For the third letter, we have 24 options, because 2 letters are already used.

For the fourth letter, we have 23 options, because 3 letters are already used.

For the fifth letter, we have 22 options, because 4 letters are already used.

Then the total number of combinations is:

Combinations = 10*9*26*25*24*23*22 = 710,424,000

So if we can not repeat digits nor letters, we can make 710,424,000 different passwords,

8 0

3 years ago