1.) 1.67×10 to the power of 9
2.) 1.133×10 to the power of -3
3.) 1.75×10 to the power of 9
4.) 2.64×10 to the power of 4
5.) 2.947×10 to the power of -6
6.) 4.24×10 to the power of 5
7.) 1.0207×10 to the power of 6
8.) 1.027×10 to the power of 5
Answer:
It is 24^21 without hesitation
There 7 blocks of hundreds which means each such block is equivalent to 100.
There are 5 blocks of tens, which means each such block is equivalent to 10.
There are 8 blocks of ones, which means each such block is equivalent to 1.
The total of these blocks will be = 7(100) + 5(10) + 8(10) = 758
We can make several two 3-digit numbers from these blocks. An example is listed below:
Example:
Using 3 hundred block, 2 tens blocks and 4 ones block to make one number and remaining blocks to make the other number. The remaining blocks will be 4 hundred blocks, 3 tens blocks and 4 ones blocks
The two numbers we will make in this case are:
1st number = 3(100) + 2(10) + 4(1) = 324
2nd number = 4(100) + 3(10) + 4(1) = 434
The sum of these two numbers is = 324 + 434 = 758
i.e. equal to the original sum of all blocks.
This way changing the number of blocks in each place value, different 3 digit numbers can be generated.
Y^3 * y^5 = y^8. you add the exponents
Answer:
The answer is "MS and QS".
Step-by-step explanation:
Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.
As NR and MQ bisect each other at S
⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ
In ΔMNS and ΔQNS
MN=QN (∵ MNQ is isosceles triangle)
∠NMS=∠NQS (∵ MNQ is isosceles triangle)
MS=SQ (Given)
By SAS rule, ΔMNS ≅ ΔQNS.
Hence, segments MS and SQ are therefore congruent by the definition of bisector.
The correct option is MS and QS