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3 years ago

Is the function increasing or decreasing over the interval (-3,-1)? Explain.

Mathematics

2 answers:

rusak2 [61]3 years ago

7 0

Increasing it is going up

Tatiana [17]3 years ago

4 0

<h3>Answer: Decreasing</h3>

Explanation:

The curve is going downhill as you move from left to right, and as you move from x = -3 to x = -1.

The interval notation of (-3,-1) means -3 < x < -1.

Since the curve is going downhill on this region, this means the function is decreasing on this interval.

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Help me pleas this is my 2 try I don't wanna fail

Vika [28.1K]

I hope this helps you

1=3

2=4

1 and 2

3 and 4 adjacent

7 0

3 years ago

In Yellowstone National Park there are 300 species of birds that migrate. This accounts for 2/7 of all the species of birds sigh

Juli2301 [7.4K]

Answer:

$\frac{2}{7} x = 300$

Step-by-step explanation:

Let x = the number of all the species of birds sighted in the park

“300 species of birds that migrate. This accounts for 2/7 of all the species of birds sighted in the park”

the above sentence means “(2/7)x = 300”

now we can solve it :

(2/7)x = 300

⇔ 2x = 7×300

⇔ 2x = 2100

⇔ x = 2100/2

⇔ x = 1050

7 0

3 years ago

Eric tossed two coins in the air at the same time. He recorded the results of each toss. He tossed the coins 100 times. About ho

Anna007 [38]

Answer:

25 times

Step-by-step explanation:

p(HT) p(hh) p(tt) p(HT)

h-heads

t-tails

p(hh)=(1/4*100)

6 0

2 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

I know it seems that alot of people like on here so people do their work but I really need help!!! I have a set of problems that

marshall27 [118]

Answer:

No Problem, I know That I personally can't stand it when people just give me the number. So another way to think of this problem is "How many 2/3 are in 3 cups." So try drawing 3 rectangles to represent 3 cups and then draw two lines inside those rectangles so that there are 3 rectangles inside each one and then shade in two of the smaller rectangles in a way that you can tell them all apart preferably different colors and then just count them.

Step-by-step explanation:

Hope this is enough to help you understand it, if not then feel free to ask for more info in the comments. :)

4 0

3 years ago