Answer:
<h3>Mass of B in Kg = -558.44kg</h3>
Step-by-step explanation:
LET'S DO THIS!
Total mass of A, B and C = 1.95kg
Mass of A = 700 kg
Mass of B = 4x the mass of c (4 x C) which is 4c
Mass of C = ? ( let's call it C )
<h3>Adding all together </h3>
700 + 4c + C = 1.95
<h3>Add like terms</h3>
700 + 5C = 1.95
5C = 1.95 - 700
5C = -698.05
C = -698.05 ÷ 5
<h3>C = -139.61</h3>
<h3>To find B now </h3><h3>Remember they said B is 4 times the mass of C and C = -139.61</h3>
therefore B = 4 × -139.61
<h3>B = -558.44 kg </h3>
<h3>To check if we are correct, we add the masses of A, B and C to see if it equals their total mass which is 1.95kg</h3>
<h3>Using your calculator: </h3>
= 700 + ( -558.44 ) + ( -139.61 )
= 700 - 558.44 -139.61
= 1.95 kg
Which makes us CORRECT ✅.
<h3>Hope this helps.</h3><h3>Good luck ✅.</h3>
Step-by-step explanation:
let's look at the full numbers under the square roots when bringing the external factors back in :
sqrt(9×9×2) - sqrt(3×3×7) + sqrt(8) - sqrt(28)
and let's present these numbers as the product of their basic prime factors
sqrt(3×3×3×3×2) - sqrt(3×3×7) + sqrt(2×2×2) - sqrt(2×2×7)
now we see that we have 2 pairs of square roots : 1 pair ends with a factor of 2, and one pair with a factor of 7.
let's combine these
sqrt (3×3×3×3×2) + sqrt(2×2×2) - sqrt(3×3×7) - sqrt (2×2×7)
and now we move the factors of 2 and 7 back out in front (of course, we need to apply the square root on these factors) :
9×sqrt(2) + 2×sqrt(2) - 3×sqrt(7) - 2×sqrt(7) =
= (9+2)×sqrt(2) - (3+2)×sqrt(7) = 11×sqrt(2) - 5×sqrt(7)
and that is the first answer option.
Answer:
(-5, -3)
Step-by-step explanation:
Scale factor 1/2 means divide points by 2
Triangularizing matrix gives the matrix that has only zeroes above or below the main diagonal. To find which option is correct we need to calculate all of them.
In all these options we calculate result and write it into row that is first mentioned:
A)R1-R3
![\left[\begin{array}{ccc}-1&0&0|0\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%260%260%7C0%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
B)2R2-R3
![\left[\begin{array}{ccc}1&0&1|1\\-2&2&1|4\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%7C1%5C%5C-2%262%261%7C4%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
C)-2R1+R3
![\left[\begin{array}{ccc}0&0&-1|-1\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%26-1%7C-1%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
D)2R1+R3
![\left[\begin{array}{ccc}4&0&3|3\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%260%263%7C3%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
E)3R1+R3
![\left[\begin{array}{ccc}5&0&4|4\\0&1&1|6\\2&0&1|1\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%260%264%7C4%5C%5C0%261%261%7C6%5C%5C2%260%261%7C1%5Cend%7Barray%7D%5Cright%5D%20)
None of the options will triangularize this matrix. The only way to <span>triangularize this matrix is
R3-2R1
</span>
![\left[\begin{array}{ccc}1&0&1|1\\0&1&1|6\\0&0&-1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%7C1%5C%5C0%261%261%7C6%5C%5C0%260%26-1%7C-1%5Cend%7Barray%7D%5Cright%5D%20%20)
<span>
This equation is similar to C) but in reverse order. Order in which rows are written is important.</span>
