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2 years ago

Which row in the chart correctly identifies the functions of structures A, B, and C? *

Mathematics

1 answer:

3241004551 [841]2 years ago

5 0

Structure A is used for (4) cell communication. The cell membrance has receptros for cell communcation.
Structure B is used to (1) extract energy from nutrients. The mitochondria is the power house of the cell.
Structure C is used for (1) protein synthesis. The golgi apparatus is where protein is created.

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Which of the following values are in the range of the function graphed below?

Elanso [62]

Hello!

The range is all of the y-values of the function. As we can see, the function is at y-values 0, -2,-4 and -6.

Therefore, our answers are 0 ,-2 and -6.

I hope this helps!

4 0

2 years ago

Read 2 more answers

A community hall is in the shape of a cuboid the hall is 40m long 15m high and 3m wide. 10 litre paint covers 25m squared costs

Darina [25.2K]

Answer:

Total cost for tiles and paints is $924.

Step-by-step explanation:

We have been given that a community hall is in the shape of a cuboid. The hall is 40m long 15m high and 3m wide.

The paint will be required for 4 walls and ceiling.

Let us find area of walls and ceiling.

$\text{Area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)$

$\text{Area of walls and ceiling}=1200+90+120$

$\text{Area of walls and ceiling}=1410$

Therefore, the area of walls and ceiling is 1410 square meters.

Given: Cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. Therefore,

$\text{ The total painting cost}=10*(\frac{1410}{25})$

$\text{ The total painting cost}=10*56.4=564$

Therefore, the total painting cost is $564.

Tiles will be required for floor. Let us find the area of floor.

$\text{Area of floor} = 40*3\text{ square meters}$

$\text{Area of floor} =120\text{ square meters}$

Given: 1m squared floor tiles costs $3. So,

$\text{Total cost for tiles} = 3*120 = 360$

Therefore, the total tiles cost is $360.

Now let us find combined total cost of tiles and paint.

$\text{Combined total cost}= 564+360 = 924$

Therefore, the combined total cost of tiles and paint is $924.

6 0

1 year ago

Given: Two concentric circles with Center Point p. How many tangent lines can be drawn that both circles share?

levacccp [35]

Answer: I think that it's not One but It's ZERO because a tangent line can only touch the circle at one point and if it touches the small circle once it would have to go through the larger one twice.

Edit: I just checked and it's definitely ZERO

7 0

1 year ago

I see that the amount i owe has been reduced from $90.00 to $75.00."

NikAS [45]

Then your amount was reduced to $15 less

4 0

2 years ago

Read 2 more answers

Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈

uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = $\{ (a,b) \in R^2 \}$

d) R1ºR4 = $\{ (a,b) \in R^2 \}$

e) R1ºR5 = R1

f) R1ºR6 = $\{ (a,b) \in R^2 \}$

g) R2ºR3 = $\{ (a,b) \in R^2 \}$

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(a,c) is in R1ºR1 if there exists b such that (a,b) is in R1 and (b,c) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2

This case is similar to the previous one. (a,c) is in R1ºR2 if there exists b such that (a,b) is in R2 and (b,c) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that ac. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0

2 years ago