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Anastasy [175]
3 years ago
9

Please help! The question is in the image.

Mathematics
1 answer:
Ghella [55]3 years ago
7 0

Answer:

my kind sir/mamn its b

Step-by-step explanation: is just muti

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Explain why the exponents cannot be added in the product 12^3 x<br> 11^3.
sergij07 [2.7K]

Answer:

see explanation

Step-by-step explanation:

The exponents can only be added if the bases are the same

Here one base is 12 and the other is 11

So the rule of addition is not applicable.

4 0
2 years ago
Read 2 more answers
How to simplify 4(2a)+7(-4b)+(3×c×5)
maw [93]
→ Solutions

⇒ Simplify <span><span><span>4<span>(<span>2a</span>)</span></span>+<span>7<span>(<span>−<span>4b</span></span>)</span></span></span>+<span><span>3c</span><span>(5)</span></span></span><span>⇒ </span><span><span><span><span><span>8a</span>+</span>−<span>28b</span></span>+<span>15<span>c

Answer
 </span></span></span></span><span>⇒ </span><span>8a−28b</span><span>+</span><span>15c</span>
5 0
3 years ago
Read 2 more answers
3+ (1/4)a = 31<br> need help !!
Olin [163]

Answer:

=

1

1

2

Step-by-step explanation:

.

3 0
3 years ago
Geometric mean. Find x, y, and z. Show work please.
Aleks [24]

Answer:

Step-by-step explanation:

i dont know the rest but the answer to x is 15 because the formula to that triangle is A^2+B^2=C^2

A=12

B=9

144(12^2) and 81 (9^2) add

225

find square root

15

z=15

4 0
3 years ago
The sums that appear when two fair? four-sided dice? (tetrahedrons) with sides 1?, 2?, 3?, and 4 are tossed
Karolina [17]

Answer:

The possible sums are 2, 3, 4, 5, 6, 7 & 8 (but they're not all equally likely)

Step-by-step explanation:

For example, the only way to get a sum of 2 is to roll a 1 and then a 1

And the only way to get a sum of 8 is to roll a 4 and then a 4 (imagine you don't roll the two dice simultaneously; this will help you to see the "other possibilities" that aren't obvious at first)

The probability of rolling a sum of 2 = probability of rolling sum of 8 (1/16 probability for each)

If you roll a 1 then a 2, or a 2 then a 1, there are 2 ways to get a sum of 3

If you roll a 3 then a 1, a 2 then a 2, or a 1 then a 3, there are 3 ways to get a sum of 4

If you roll a 4 then a 1, a 3 then a 2, a 2 then a 3, or a 1 then a 4, there are 4 ways to get a sum of 5 (5 is the most likely sum)

If you roll a 4 then a 2, a 3 then a 3, or a 2 then a 4, there are 3 ways to get a sum of 6

If you roll a 4 then a 3, or a 3 then a 4, there are 2 ways to get a sum of 7

5 0
3 years ago
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