Answer:
From left to right,
1. impossible
2. unlikely
3. neither unlikely nor likely
4. likely
5. certain
Step-by-step explanation:
Use the area of rectangle formula which is:
Given that:
- area of rectangle = 16
- width = 8
- length = x-5
Substitute in the formula:
Distribute 8 in the expression:
Move -40 to another side and change from -40 to 40.
Then move 8 to divide 56, leaving only x.
Hence, the answer is:
No they won’t be.Consider the linear combination (1)(u – v) + (1) (v – w) + (-1)(u – w).This will add to 0. But the coefficients aren’t all 0.Therefore, those vectors aren’t linearly independent.
You can try an example of this with (1, 0, 0), (0, 1, 0), and (0, 0, 1), the usual basis vectors of R3.
That method relied on spotting the solution immediately.If you couldn’t see that, then there’s another approach to the problem.
We know that u, v, w are linearly independent vectors.So if au + bv + cw = 0, then a, b, and c are all 0 by definition.
Suppose we wanted to ask whether u – v, v – w, and u – w are linearly independent.Then we’d like to see if there are non-zero coefficients in the linear combinationd(u – v) + e(v – w) + f(u – w) = 0, where d, e, and f are scalars.
Distributing, we get du – dv + ev – ew + fu – fw = 0.Then regrouping by vector: (d + f)u + (-d +e)v + (-e – f)w = 0.
But now we have a linear combo of u, v, and w vectors.Therefore, all the coefficients must be 0.So d + f = 0, -d + e = 0, and –e – f = 0. It turns out that there’s a free variable in this solution.Say you let d be the free variable.Then we see f = -d and e = d.
Then any solution of the form (d, e, f) = (d, d, -d) will make (d + f)u + (-d +e)v + (-e – f)w = 0 a true statement.
Let d = 1 and you get our original solution. You can let d = 2, 3, or anything if you want.
1>>g
2>>e
3>>h
4>>f
5>>b
6>j
7>>k
8>>c
9>>I
10>>d
11>>a
;)
Answer:
y = (-2/5)x - 2
Step-by-step explanation:
One way to attack this problem is to interchange the coefficients of x and y and change the sign of one to +: 5x - 2y = -6 becomes 2x + 5y = c. Solving for the slope, m, we get 5y = -2x + c first, and then y = (-2/5)x + D.
Subbing 5 for x and -4 for y, we now have -4 = (-2/5)(5) + D.
Then -4 = -2 + D, so that D = -2.
The desired equation is thus y = (-2/5)x - 2.
Check: Does this pass through (5, -4)? Is -4 = (-2/5)(5) - 2 true? Yes.
Is the slope -2/5 the negative reciprocal of 5/2? Yes, it is.
