Answer:
Step-by-step explanation:
4A Area is 40. The area is the length times the width giving you an answer of 40 sq in.
4A Perimeter is is 2L+2W. Multiply the length by 2 then the width by two and add them together this gives you 8+20 which is 28 inches.
Answer:
17
Step-by-step explanation:
Second square number is 9; 3×3
second cube number is 8; 2×2×2
9+8=17.
Answer:
35.34 centimeters
Step-by-step explanation:
The amount of pizza is same as the area. The pizza is in a shape of a circle. The area of a circle is:
Where
r is the radius (half of diameter)
Area of one medium size pizza is (diameter 25 means radius is 25/2 = 12.5):
So, area of two of these medium size pizzas is:
490.87 * 2 = 981.74
This should be the area of the large pizza. We can find the radius of that by substituting into formula. Shown below:
Hence, the diameter would be around:
17.67 * 2 = 35.34 centimeters
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.
