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jasenka [17]
3 years ago
6

One uranium atom has a diameter of 3.5x10^-8 centimeters. What is the sum of 10,000 uranium atoms'

Mathematics
1 answer:
Vladimir [108]3 years ago
6 0

Answer:

3.5 * 10^-4 centimeters

Step-by-step explanation:

Here, we have one uranium atom having a diameter of 3.5 * 10^-8 centimeters.

We now want to find the sum of 10,000 uranium atoms diameters

Mathematically;

That would be ;

3.5 * 10^-8 * 10,000

But 10,000 = 10^4

Thus, we have

3.5 * 10^-8 * 10^4

= 3.5 * 10^-4 centimeters

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Explain your answer!! <br><br> Will give brainlst
Novosadov [1.4K]
• So we know that.....

x represent bags of snack and y is bottles of water.


This equations shows the total amount and the cost of each water bottle and snack:

20.00 = 2.50x + 1.00y

Total: $20.00
Snack: $2.50
Water Bottle: $1.00


And this question shows the total items:

11 = x + y

Which there will be some snack + some water bottle = 11 items


—————————————————————

• Now I’m going to first solve for x, which is the amount of bags of snack.

I will use the equation, 11 = x + y.

(First, we’ll subtract y from both side, since we’re solving for x [UNDO])

11 = x + y
-y = - y
_______
11 - y = x —> so x is equal to 11 minus y.


—————————————————————

• Now we’re going to plug the 11 - y as x in the equation: 20.00 = 2.50x + 1.00y to solve for y.

20.00 = 2.50 (11 - y) + 1.00y

20.00 = 27.5 - 2.50y + 1.00y (Distributed)

20.00 = 27.5 - 1.50y (Combine like terms)

20.00 = 27.5 - 1.50y
-27.5 = -27.5 (Subtract -27.5 both side)
——————————
-7.5 = - 1.50y


-7.5 = -1.50y
—— ——— (Divide both side by -1.50)
- 1.50 = -1.50

5 = y

y is equals to 5, which means that there are 5 water bottles.


Now we know there are 11 items total and because there are 5 water bottles, there will be 6 bags of snacks. 11-5=6

—————————————————————

ANSWER:

They bought 6 bags of snacks! :)
4 0
3 years ago
If y=34 what is y50+y/60
zmey [24]
34 times 50 plus 34 divided by 60 equals 28.9
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sasho [114]

Answer:

what is the options

Step-by-step explanation:

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3 years ago
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Which additional information could be used to prove that the triangles are congruent using AAS or ASA? Check all that apply.
Ad libitum [116K]

Consider the given triangles.

Given: \angle C \cong \angle Q

ASA congruence criterion states that if two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

AAS congruence criterion states that if two angles and a non-included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

Consider the first part:

1. In triangles ABC and QPT

If we take \angle B \cong \angle P , BC \cong PQ along with the given condition.

Then the triangles are congruent by ASA congruence criterion.

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Then the triangles are congruent by AAS congruence criterion.

5. If we take AC=TQ=3.2 and CB=QP=3.2 along with the given condition, then the triangles are congruent but by SAS congruence criteria neither by ASA nor AAS congruence criterion.

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3 years ago
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alekssr [168]
Total questions are 30.

70% of 30
=30(70/100)
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So, the questions can be wrong are 30-21.
=9 questions..

Thanks!
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3 years ago
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