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2 years ago

13

You were charged $51.50 last month for data usage on your smartphone. Your data plan charges a monthly fee of $20 plus $2.50 for

each extra gigabyte (GB) used. How many extra gigabytes did you use last month?

2 answers:

eimsori [14]2 years ago

6 0

He went 13 gigabytes over his limit
HOPES DIS HELPS

KIM [24]2 years ago

5 0

20 + 2.5g = 51.50
2.5g = 51.50 - 20
2.5g = 31.50
g = 31.50 / 2.5
g = 12.6 <=== there were 12.6 extra gigabites used

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Strike441 [17]

If the first two statements are true, then the third statement:

"Bandit is not a dog"

Is true.

<h3>if the first two statements are true, is the third statement true?</h3>

The first statement is:

P = "All dogs like walks"

We assume that this is true, so, if X is a dog, X likes walks.

The second statement is:

Q = "Bandit doesn't like walks"

From this, we only can conclude that Bandit doesn't like walks.

And all dogs like walks, so if Bandit were a dog, he would like walks, but because he does not like walks, we can conclude that Bandit is not a dog.

From that, the third statement:

K = "Bandit is not a dog"

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So we write that as a conditional statement:

If P and Q ⇒ K

<em>"if all dogs like walks, and Bandit doesn't like walks, then bandit is not a dog".</em>

If you want to learn more about conditional statements:

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1 year ago

The elevation E, in meters, above sea level at which the boiling point of a certain liquid is t degrees Celsius is given by the

Anna007 [38]

Answer:

$E(119.5)= 730m$

Step-by-step explanation:

Given

$E(t)= 1200(120-t)+520(120-t)^2$

Required

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$E(t)= 1200(120-t)+520(120-t)^2$

Simply substitute 119.5 for t

$E(119.5)= 1200(120-119.5)+520(120-119.5)^2$

Evaluate the expressions in bracket

$E(119.5)= 1200(0.5)+520(0.5)^2$

Solve 0.5²

$E(119.5)= 1200 * 0.5+520*0.25$

$E(119.5)= 600+130$

$E(119.5)= 730$

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3 years ago

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

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Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

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$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

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Anettt [7]

The answer is more than thirty

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lozanna [386]

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