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LUCKY_DIMON [66]
3 years ago
11

Which one is the better buy $3 and 2 refills $4 and 3 refills

Mathematics
2 answers:
Brums [2.3K]3 years ago
6 0

Answer:


Step-by-step explanation:

3 and 2

Vikki [24]3 years ago
6 0

Well, for the first one, you would be paying 0.66 cents for each refill.

For the second one, you would be paying 0.75 cents for each refill.

So, the first option is better because it is cheaper.

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Please I need this ​
Free_Kalibri [48]

Answer:

1) not

2) QE

3) QE

4) QE

5) QE

6) QE

7) QE

8) not

9) QE

10) not

hope this helps you!

7 0
3 years ago
If you can answer this and show your work ill give brainly
Arlecino [84]

Answer:

First answer is 127 and second answer is 15

8 0
2 years ago
Read 2 more answers
Rewrite 0.37 in expanded form using powers of ten.
Black_prince [1.1K]

Expanded form is the form of writing numbers in addition. Example: 653 would be 600+50+3 in expanded form. .37 using powers would be 37×10^-2 .

In expanded form, we just need to expand the 37 because it is the only one with 2 different places without a 0 in either one.

Answer: 30+7×10^-2 would be the expanded form of .37 using powers of 10.

4 0
3 years ago
I WILL MARK BRAINLIEST!!! !!<br><br><br> (picture included with ONE question ^^^^^) only part a
Taya2010 [7]

Answer:

A. y = 1.5m + 7

B. at 7 months the baby will weigh 17.5 pounds

C. Since the baby is born weighing 7 pounds, at 0 months he will weigh 7 pounds. This tells us that the y-intercept will be seven. And since the baby is gaining 1.5 pounds every month, this means that the slope will be 1.5 and m will be the x of our y = mx + b equation. For part B, we simply plug in 7 to m and solve by multiplying 1.5 and 7 to get 10.5, then adding 7 to get 17.5.

6 0
3 years ago
Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges
sleet_krkn [62]

Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

f(x)=a^{x}\\ f'(x)=a^xln(a)

The process will help us to understand what is happening, at first we rewrite the function:

f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\

And then, we use the chain rule to differentiate:

f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know  that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b)  (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

3 0
3 years ago
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