Answer: √58 cm or 7.62 cm
Step-by-step explanation:
The Pythagorean Theorem is a²+b²=c². Since we are given the lengths of a and b, we can plug them into this formula to find the length of the hypotenuse.
7²+3²=c²
49+9=c²
58=c²
c=√58 or 7.62 cm
1:3 :) why? well if we think about it 2:9 is just not gonna work, cross that out. 3:18? no, thanks. 1:3 is it
Well 32 x 405 is 12,960, but it ask for 2 numbers added to make 12,960 so the answer is 810 + 12,150 which is 2 products that add up to the answer 12,960.
Answer:
The number of once is 9.1
The number of hundreds is 8.9
Step-by-step explanation:
Given as :
The total of digits having ones and hundreds = 900
The sum of digits = 18
Let The number of ones digit = O
And The number of hundreds digit = H
So, According to question
H + O = 18 .........1
100 × H + 1 × O = 900 ........2
Solving the equation
( 100 × H - H ) + ( O - O ) = 900 - 18
Or, 99 H + 0 = 882
Or , 99 H = 882
∴ H = 
I.e H = 8.9
Put the value of H in eq 1
So, O = 18 - H
I.e O = 18 - 8.9
∴ O = 9.1
So, number of once = 9.1
number of hundreds = 8.9
Hence The number of once is 9.1 and The number of hundreds is 8.9
Answer
Answer:
the lower right matrix is the third correct choice
Step-by-step explanation:
Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).
Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.
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We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.
The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...
![\left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%261%2629000%5C%5C1%2B2%261-3%261-1%261000%2829%2B1%29%5C%5C0%260.15%260.15%262100%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%261%2629000%5C%5C3%26-2%260%2630000%5C%5C0%260.15%260.15%262100%5Cend%7Barray%7D%5Cright%5D)
Already, we see that the second row matches that in the lower right matrix.
The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.
The correct choices here are the two you have selected, and <em>the lower right matrix</em>.