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valentinak56 [21]
3 years ago
5

I need help quick!!!!! Ive been trying to figure this out for 20 mins now.

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
3 0

Answer:

take one finger and than the other and how may fingers do you have now 2

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What are the coordinates of this and radius!
LiRa [457]

-1,1 I already told u, nut

3 0
3 years ago
-3.75+2(-4x+6.1)-3.25
Pani-rosa [81]

The value of x is 0.65

<u>Step-by-step explanation:</u>

This solution is given by solving the given problem using BODMAS rule.

This means that the steps to be followed to find the solution should be in the order of BODMAS rule.

B- Bracket, O- of, D- Division, M- Multiplication, A- Addition, S- Subtraction.

This rule states that the 1st operation must be done for Brackets.

⇒ -3.75+2(-4x+6.1)-3.25 = 0

⇒ -3.75 -8x + 12.2 -3.25 = 0

The addition operation should be performed next and then followed by subtraction.

⇒ -8x -7+ 12.2 = 0

<u>To find the value of x :</u>

Keeping the x term alone on one side and moving the constants on other side,

⇒ -8x = -12.2 + 7

⇒ -8x = -5.2

⇒ x = 5.2/8

⇒ x = 0.65

3 0
3 years ago
A train traveled the first 200 miles of its trip at a speed of 50 mph and the next 130 miles at a speed of 65 mph. What was the
12345 [234]
55 miles per hour, to answer questions like these you need to calculate the time (T=D/R) for each individual portion of the trip, add the time together to get total time, and then divide the total distance by the total time in order to get the average velocity in mph.
4 0
3 years ago
P=?,r=5%,t=13 months, i=$103.55
V125BC [204]

Answer:

$1917.60

Step-by-step explanation:

Interest = Principal x Rate x Time

103.55 = (.05)(13/12)P

103.55 = .054P

P = 1917.60

8 0
3 years ago
assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28
melomori [17]

Given:

x and y are both differentiable functions of t.

4x^2+3y^3=28

x=-1\text{ and }\dfrac{dy}{dt}=8

To find:

The value of \dfrac{dx}{dt}.

Solution:

We have,

4x^2+3y^3=28       ...(i)

At x=-1,

4(-1)^2+3y^3=28

4+3y^3=28

3y^3=28-4

3y^3=24

Divide both sides by 3.

y^3=8

Taking cube root on both sides.

y=2

So, y=2 at x=-1.

Differentiate (i) with respect to t.

4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0

Putting x=-1, y=2 and \dfrac{dy}{dt}=8, we get

4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0

-8\dfrac{dx}{dt}+9(4)(8)=0

-8(\dfrac{dx}{dt}-9(4))=0

Divide both sides by -8.

\dfrac{dx}{dt}-36=0

\dfrac{dx}{dt}=36

Therefore, the value of 4x^2+3y^3=28 is 36.

6 0
2 years ago
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