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Ostrovityanka [42]
2 years ago
5

A recycling station brought in 1/2 T of newspapers on Monday. On Tuesday, the station collected only half of what it did on Mond

ay. On Wednesday the station collected only half of what it did on Tuesday. How many pounds did it collect in total during the three days?
Mathematics
1 answer:
Oxana [17]2 years ago
8 0
Its b took the test Rita b duemwowms
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3pt and 2c to ? oz <br><br><br>help asap plzzzzzzzz
brilliants [131]

Answer:

What u said dosent make sense lol

8 0
3 years ago
Read 2 more answers
What is the height of a cylinder with a volume of 1205.76 cubic inches and a radius of 8 inches? Round to the nearest tenth of a
Anit [1.1K]
Cylinder height = volume / PI * radius^2
cylinder height =1,205.76 / PI * 8^2
cylinder height = <span> <span> <span> 5.9969582557 </span> </span> </span>
which I guess can be safely rounded to 6 inches
Source www.1728.com.diamform.htm


3 0
3 years ago
Help you smart people
Vinil7 [7]
35 ≥ 12.50+1.50g
this is because cole can’t go over the amount $35.

Step 1: Simplify both sides of the inequality.
35≥1.5g+12.5

Step 2: Flip the equation.
1.5g+12.5≤35

Step 3: Subtract 12.5 from both sides.
1.5g+12.5−12.5≤35−12.5
1.5g≤ 22.5

Step 4: Divide both sides by 1.5.
1.5g 22.5
1.5 ≤ 1.5g

g≤15

hope this helps :)

8 0
3 years ago
(Prove) The angle subtended by an arc at the center is double the angle subtended by it at any
Pavlova-9 [17]

Answer:

The angle subtended by an arc at the center is double the angle subtended by it at any  point on the remaining part of the circle.

Step-by-step explanation:

Let us consider the image attached.

Center of circle be O.

Arc AB subtends the angle \angle APB on the circle and \angle AOB on the center of the circle.

To prove:

\angle AOB = 2 \times \angle APB

Proof:

In \triangle PAO: AO and PO are radius of the circles so AO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, \angle PAO = \angle OPA

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

\angle AOQ = \angle PAO + \angle OPA=2 \times \angle APO .... (1)

Similarly,

In \triangle PBO: BO and PO are radius of the circles so BO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, \angle PBO = \angle OPB

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

\angle BOQ = \angle PBO + \angle OPB=2 \times \angle BPO .... (2)

Now, we can see that:

\angle AOB = \angle AOQ+\angle BOQ

Using equations (1) and (2):

\angle AOB = 2\angle APO+2\angle BPO\\\angle AOB = 2(\angle APO+\angle BPO)\\\bold{\angle AOB = 2(\angle APB)}

Hence, proved.

4 0
3 years ago
A credit card had an APR of 25.67% all of last year and compounded interest daily. What was the credit card's effective interest
dimulka [17.4K]

Answer:

The answer is 29.25%.



7 0
3 years ago
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