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sergey [27]
2 years ago
5

Between 20 to 35 degrees north latitude, and also between 20 to 35 degrees south latitude are found:

Mathematics
1 answer:
myrzilka [38]2 years ago
5 0

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Its area includes all Sahara's locations in North Africa, South Arabia, Iran's and Iraq's larger parts, North-Western India, California throughout the United States, South Africa but most of Australia.

Half-arid temperatures include places like the Utah, Montana, and Gulf Coastal regions of sagebrush. Also, it comprises regions in Iceland, Russia, Scandinavia, Greenland, and Northeast India. Semi-arid thankless than tube called per year of rain and up to 20 inches per year at much more than arid deserts.

Regions from of the latitude of 25° to 35° usually develop desert, because air sinks and is heated under pressures in this area. The world's dry and semi-arid desert regions are between 20°C and 35°C north latitude and between 20°C and 35°C South latitude.

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The figure below shows the graph of f ', the derivative of the function f, on the closed interval from x = -2 to x = 6. The grap
wolverine [178]

Answer:

x = -2

Step-by-step explanation:

From x = -2 to x = 5, f' is negative.  That means f is decreasing.

From x = 5 to x = 6, f' is positive.  That means f is increasing.

The negative area (between x = -2 and x = 5) is larger than the positive area area (between x = 5 and x = 6).  That means f decreases more than it increases.

So f is an absolute maximum at x = -2.

6 0
2 years ago
Leah decided to paint some of the rooms at her 51-room hotel. She needs 1/5 of a can of paint per room. If Leah had 6 cans of pa
netineya [11]

Answer:

Leah can paint 30 rooms.

Step-by-step explanation:

Each room requires \dfrac{1}{5} of a can of paint, this means painting 5 rooms requires

\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}=\dfrac{1+1+1+1+1}{5}=\dfrac{5}{5} =1

or 1 can paints 5 rooms.

So, if Leah has 6 cans of paint, then how many rooms can she paint?

Leah can paint

\dfrac{5\:rooms}{can} *6\:cans=\boxed{30\: rooms}

Leah can paint 30 rooms with 6 cans of paint.

5 0
2 years ago
Function f is a cosine function with period 8π, amplitude 3, and a local maximum at f(0) = –3. What is the equation of the midli
alisha [4.7K]

Answer:

y = 0

Step-by-step explanation:

Since there is no indication of a shift in the graph of f(x) = 3cos(-0.25x), then the midline must be the normal midline for a cosine or sine function which is: y = 0.

8 0
2 years ago
Read 2 more answers
Harry has 20 sweets. He gives 7 of the sweets to Nadia. What fraction of the 20 sweets does Harry have now?
olganol [36]

Answer: 13

Step-by-step explanation:

1. Harry has 20/20 sweets

2. Nadia takes 7/20 sweets

3. Your left with 13/20

3 0
2 years ago
Read 2 more answers
Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.
xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form p(t)=at^2, where a is in ℝ.

2.Let W2 be the set of all polynomials of the form p(t)=t^2+a, where a is in ℝ.

3. Let W3 be the set of all polynomials of the form p(t)=at^2+at, where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

p_1 (t) = at^2, p_2(t)=bt^2.

We must check that p_1+p_2(t) is in W1.

Note that

p_1(t)+p_2(t) = at^2+bt^2  = (a+b)t^2

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say p_1(t) = a(t^2+t), p_2(t)= b(t^2+t). Then

p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 p(t) = at^2and a real scalar b. Then

bp(t) = b(at^2) = (ba)t^2).

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 p(t) = a(t^2+t)and a real scalar b. Then

bp(t) = b(a(t^2+t)) = (ba)(t^2+t).

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.  

6 0
2 years ago
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