Round 10.054 To 2 decimal places

Answers

Answer 1
Answer: the answer to "Round 10.054 To 2 decimal places" is 10.05

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What is the LCD of 2/7 and 9/10

Answers

Assuming LCD means Lowest Common Denominator
Assuming by "and" you mean adding the two

Well, you go by the multiples of 7;
7, 14, 21, 28, 35, 42, 49, 56, 63, 70
Multiples of 10;
10, 20, 30, 40, 50, 60, 70, 80, 90, 100

Straight away you can see the lowest common multiple is 70,

Now to go about adding them,
2/7 * 10 (both sides) , 20/70
9/10 * 7 (both sides) , 63/70

Then you add the two; 20/70 + 63/70  - now we can add them because they share the same denominator
20/70 + 63/70 = 83/70 = 1 whole 13/70 (simplified)

The LCD is 70

Quick tip, 
to find a common multiple (not necessarily the lowest) multiply the two denominators. 
The least common denominator or LCD is 70 because for
10=10,20,30,40,50,60,70,80
7=7,14,21,28,35,42,49,56,63,70,77
and 70 is the least common denominator of 10 and 7 so 70 ims your answer

Lily has 3 dog toys that are red one fourth of all her dog toys are red how many dog toys lily have

Answers

3=red
1/4 times all=red
1/4 times all=3
times 4 both sides
all=12
She has 12 dog toys in all 3 x 4 = 12

How to graph 3x+1=3x+2

Answers

That equation has no graph. If it had a solution, the solution would be a single number. But that particular equation has no solution either ... there is no number that 'x' could be that would make the equation true.

Joe has some pocket money he spends 50 percent of it he then spends another 1.50 he has 75p left how much pocket money did he have

Answers

2.25, because you have to add the money percentage he spent to find what he had at first.

jayson buys a car and pays by installments.each installmentsis $567 per month.after 48 month, jayson owes $1250. what was the total price of the vehicle ?

Answers

Let

T--------> the total price of the vehicle

x--------> amount paid

y--------> amount owed

we know that

the total cost of the vehicle is equal to the amount paid plus the amount owed

so

T=x+y

Find the amount paid

x=567*48\n x=27,216\ dollars

Find the amount owed

y=1,250\ dollars

Find the total cost of the vehicle

T=x+y\n T=27,216+1,250\n T=28,466\ dollars

therefore

the answer is

the total price of the vehicle is 28,466\ dollars

The total price of the vehicle is 28,466

  • Each installment monthly = $567
  • Amount owes by Jayson after 48 months = $1250

Further Explanation

To determine the total price of the vehicle, we can solve this question using the area model (please refer to the attachment below)

Firstly, let’s use the area model method to calculate the total price of the vehicle.

Therefore, we have:

567 x 48

= (500 + 60 + 7) x (40 + 8)

= (500 x 40) + (500 x 8) + (60 x 40) + (60 x 8) + (7 x 40) + (7 x 8)

= 20000 + 4000 + 2400 + 480 + 280 + 56

= 27216.

Recall, Jaycee still owes $1250 after 48 months; therefore we can derive the total price of the vehicle by adding the total money for the whole 48 months and the amount Jayson owes.

Therefore we have:

27216 + 1250

= 28466

Thus, the total price of the vehicle is $28,466

The area model refers to a rectangular diagram used in solving multiplication and division problems in mathematics.

LEARN MORE:

KEYWORDS:

  • jayson
  • area models
  • vehicle
  • $567 per month
  • jayson owes $1250
  • the total price of the vehicle

The graph of which function has a minimum located at (4, –3)?f(x) = x2 + 4x – 11
f(x) = –2x2 + 16x – 35
f(x) = x2 – 4x + 5
f(x) = 2x2 – 16x + 35

Answers

Answer:

Step-by-step explanation:

In order to solve the question, we have to derivate each function.

1) f(x) = x2 +4x -11

Then,

f'(x)= 2x +4

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 2*4 +4 = 12 ≠ 0 then this function doesn't not have a minimum at (4, -3)

2) f(x) = –2x2 + 16x – 35

Then,

f'(x)= -4x +16

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= -4*4 16 = 4 ≠ 0 then this function have a critical point at (4, -3)

then,

f''(4) =-4 <0 then we have a minimum at (4, -3)

3) f(x) = x2 – 4x + 5

Then,

f'(x)= 2x -4

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 2*4 -4 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)

4) f(x) =  2x2 – 16x + 35

Then,

f'(x)= 4x -16

Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:

f'(4)= 4*4 -16 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)

It is the third option on Edge