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BabaBlast [244]
3 years ago
6

Every week, Mr. Kirkson uses 316 gallon of water to water every 13 square foot of his garden.How many gallons of water does Mr.

Kirkson use per square foot to water his garden each week?
Enter your answer in the box as a fraction in simplest form.
Mathematics
2 answers:
nydimaria [60]3 years ago
7 0
24 4/13 gallons are used to water one square foot of his garden.
charle [14.2K]3 years ago
5 0

if this is supposed to be 3/16 and 1/3 the right answer is 9/16. I took the test :)

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Solve the inequality. -x ≤ 5​
yanalaym [24]
The answer is x ≥ −5
6 0
3 years ago
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A dog runs 100 meters in 30 seconds. How fast is he running
matrenka [14]

Answer:

3.33 m/s

Step-by-step explanation:

Distance divided by time = speed

100/30 = 3.33 m/s

6 0
2 years ago
What is -2 × 1 1/4=
RoseWind [281]

Answer: -11/2

Step-by-step explanation:

Just set it up so it looks like this:

-2/1 x 11/4 and multiply across the top, which would be -22 and then the across the bottom, which would equal 4. So it’s -22/4. BUT you can simplify it down to -11/2 by dividing both sides of the fraction by 2. So then answer is -11/2

Hope that makes sense! :)

5 0
2 years ago
If f(x)=| x-2 |, then find f(-3)
GenaCL600 [577]

The value of f(-3) is 5.

In order to find this answer, we are going to input -3 in for x (this is what the notation f(-3) means).

f(x)=| x-2 |

f(-3)=| -3 -2 |

f(-3)=| -5 |

Now we have to take the absolute value.

f(-3) = 5

3 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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