The four hand of a clock moves from 12 to 5 o'clock through how many degrees does it move?

Answers

Answer 1
Answer: to find the answer you have to first find how many degrees are there in between each number in the clock. the clock is 360°degrees and there are 12 numbers marked in a clock., so you have to divide 360 by 12 to get the number of degrees in between each number,
360°÷12=30° so there are 30°degrees between each number. 

if a clock hand moves from 12 to 5, it pass 5 numbers, so to get your final answer you have to multiply 30° by 5=150°.

so your answer is 150°degrees.
hope you can understand what i said. :)

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What is the steps to solve 25 milligrams per day for one week

Answers


I don't really know what the question is that needs to be solved, but
here's a suggestion that might help:

Since there are seven days in a week, you could multiply 25 milligrams
by 7, and that will tell you the number of milligrams in a whole week.

If 5,x,y,40 are in geometrical progression, find x and y respectively.

Answers

Answer:

x = 10, y = 20

Step-by-step explanation:

Because this is a geometric sequence, we'll call the rate of change z. Because 40 is 3 terms away from 5, we can write 5 * z * z * z = 5z³ = 40.

z³ = 8 → z = 2

Now, we simply multiply 5 by 2 to get x, which is 10, and then we multiply x by 2 to get 10 * 2 = 20 for y. Hope this helps!

Final answer:

To find the values of x and y in the given geometric progression, we use the concept of a common ratio. By setting up equations based on the definition of a geometric progression, we can solve for x and y. In this case, x is 10 and y is 20.

Explanation:

To find the values of x and y in the given geometric progression, we need to identify the common ratio between the terms. In a geometric progression, each term is obtained by multiplying the previous term by the common ratio. Therefore, we have:




  1.  
  2. 5 * r = x

  3.  
  4. x * r = y

  5.  
  6. y * r = 40



From the first equation, we can substitute x as 5 * r in the second equation:



5 * r * r = y



Then, we substitute y as 5 * r * r in the third equation:



5 * r * r * r = 40



Simplifying the equation, we get:



r^3 = 40/5 = 8



Taking the cube root of both sides:



r = 2



Substituting this value of r back into the equations, we find that x = 10 and y = 20.

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A coordinate grid with 2 lines. The first line labeled f(x) passes through (negative 3, 3), (0, 2), and (3, 1). The second line labeled g(x) passes through the points at (negative 3, 0) and (0, 2) What is the solution to the system of linear equations? (–3, 0) (–3, 3) (0, 2) (3, 1)

Answers

The solution to the system of linear equations is (0,2). The correct option is C.

What is a system of linear equations?

A finite set of equations for which common solutions are sought is referred to in mathematics as a set of simultaneous equations, often known as a system of equations or an equation system.

It is given that in a coordinate grid with 2 lines. The first line labelled f(x) passes through (-3, 3), (0,2), and (3, 1). The second line labelled g(x)passes through the points at (-3, 0) and (0, 2).

Plotting the two lines by using the points on the graph. After plotting the points there will be two lines that will intersect at the point ( 0, 2 ). Then the solution of the equation is ( 0,2).

Therefore,the solution to the system of linear equations is (0,2). The correct option is C.

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Answer:

(0, 2) or C

Step-by-step explanation:

What two whole numbers is the quotient between? 38\div438÷438, divided by, 4 Choose 1 answer: Choose 1 answer: (Choice A) A 333 and 444 (Choice B) B 555 and 666 (Choice C) C 888 and 999 (Choice D) D 999 and 101010

Answers

Answer:

(D) 9 and 10

Step-by-step explanation:

Given the quotient: 38÷4

(38)/(4)=9 (2)/(4)=9 (1)/(2)

Since 9 (1)/(2) is greater than 9 but less than 10,

9<9 (1)/(2)<10

It therefore lies in between 9 and 10.

Given that (-9,-3) is on the graph of f(x), find the corresponding point f(x+1)

Answers

the answer is -8
because f(x+1) where x= -9
-9+1 = -8

Explain how unwrapping a present is used as an analogy for solving an equation. How do you "unwrap" with an equation?...?

Answers

Here is an analogy comparing between unwrapping a present and unwrapping an equation. Both actions do require following a process. You cannot just skip one step in order to arrive at the answer. Just like unwrapping a present, you need to do it in order. First, you take off the ribbon, then you start to unwrap the wrapper until you see the actual gift. On the other hand, when you are unwrapping an equation, you also follow a certain step. You are given a certain formula and values. What you need to do next is to substitute, do the solution, until it is simplified. And lastly, there you have it. You can now see the actual answer. Hope this answer helps.

Final answer:

Solving an equation is likened to unwrapping a present because in both cases, you aim to uncover what's within by dealing with the outer layers systematically. Through an inverse process, you isolate the equation's variable and solve for it, just as the step-by-step unwrapping of layers reveals the gift within the package.

Explanation:

Unwrapping a present is a great way to understand the process of solving an equation. If you consider the equation itself as being 'wrapped' in different operations such as multiplication, division, addition, and subtraction, you can begin to 'unwrap' it by systematically dealing with these operations in reverse from the order of operations. This would be like when you unpack the present by undoing the tape, then the paper, then the box, or whatever other layers might be present.

To use a direct example, let's consider the equation 3x + 7 = 22. You start unwrapping by isolating for the term with the variable. This means you handle the operation that does not involve the variable first. So, subtract 7 from both sides of the equation to get 3x = 15. Next, you handle the multiplication operation by dividing both sides by 3 and solving for x, which would be 5 in this case. Just like unwrapping a present, we aim to get to the 'inside' or the root of the equation.

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