$\left[\begin{array}{ccc}-1&3\\5&4\end{array}\right] \cdot \left[\begin{array}{ccc}0&-2&6\\1&-3&2\end{array}\right]\\\\= \left[\begin{array}{ccc}(-1)(0)+(3)(1)&(-1)(-2)+(3)(-3)&(-1)(6)+(3)(2)\\(5)(0)+(4)(1)&(5)(-2)+(4)(-3)&(5)(6)+(4)(2)\end{array}\right]\\\\= \left[\begin{array}{ccc}0+3&2-9&-6+6\\0+4&-10-12&30+8\end{array}\right]\\\\ =\left[\begin{array}{ccc}3&-7&0\\4&-22&38\end{array}\right]$

3 0

3 years ago

1/3 radical 176xy to the 8th power

Tanya [424]

Answer:

Remove all perfect squares from inside the square root. ... I think it's about eighth or ninth grade. ... so if you have the cube root of the square root of (x-5) =2, you get ((x-5)^(1/2))^1/3 = 2, power to power requires multiplication, so (x-5)^1/6 = 2, ...

Missing: 176 ‎xy

5 0

3 years ago

The sum of the first 15 terms of a geometric sequence with 7 as the first term and a common ratio of -3 is __________.

Andre45 [30]

Step-by-step explanation:

the formula for the sum of the first n terms of a geometric sequence is

Sn = s1(1 - r^n)/(1 - r)

with r being the common ratio and s1 is the first term.

so,

S15 = 7×(1 - (-3)¹⁵)/(1 - -3) = 7×(1 - -14,348,907)/4 =

= 7×14,348,908/4 = 7×3,587,227 = 25,110,589

4 0

2 years ago