Answer:
125.4
Step-by-step explanation:
Given
Required
Round to 1 decimal place
Up till the first decimal place, the number is:
The digit after .3 is 5
The conditions for approximation are:
- If n > 4, approximate to 1
In this case: 5 > 4, so we approximate to 1
Add this "1" to the last digit of 125.3. This becomes 125.4
<em>Hence: when the number is approximated to 1 decimal place, the digit is 125.4</em>
100 times 17 bc u have to count the shdded parts
0.17 because its in the hundredths place
Answer:
2.9
Step-by-step explanation:
Plz mark as brainliest!!!
Using the determinant method, the cross product is
so the answer is B.
Or you can apply the properties of the cross product. By distributivity, we have
(3i + 8j - 6k) x (-4i - 2j - 3k)
= -12(i x i) - 32(j x i) + 24(k x i) - 6(i x j) - 16(j x j) + 12(k x j) - 9(i x k) - 24(j x k) + 18(k x k)
Now recall that
- (i x i) = (j x j) = (k x k) = 0 (the zero vector)
- (i x j) = k
- (j x k) = i
- (k x i) = j
- (a x b) = -(b x a) for any two vectors a and b
Putting these rules together, we get
(3i + 8j - 6k) x (-4i - 2j - 3k)
= -32(-k) + 24j - 6k + 12(-i) - 9(-j) - 24i
= (-12 - 24)i + (24 + 9)j + (32 - 6)k
= -36i + 33j + 26k
