Answer: The difference in temperature is -44 degrees.
Explanation: Since the new temperature is a negative number and the old was positive, you add them together. However instead of (-5) +39 (which the answer would be 35), you would write 5 +39 because -5 is 5 spaces away from 0. You can also check by using the equation 39 -44. You would get the answer -5, proving that the answer is indeed -44 degrees.
Answer:
Hence the area of each of the triangles are:
49 cm^2 and 16 cm^2.
Step-by-step explanation:
If the two triangles are similar; the perimeters of the two triangles are in the same ratio as the sides.
We are given that the ratio of the perimeters of two similar triangles is 4:7.
Let a and b denotes perimeter of two triangles.
i.e. here a:b=4:7.
Let A and B denote the area of two triangles.
also we know that for two similar triangles;
![\dfrac{A}{B}=\dfrac{a^2}{b^2}](https://tex.z-dn.net/?f=%5Cdfrac%7BA%7D%7BB%7D%3D%5Cdfrac%7Ba%5E2%7D%7Bb%5E2%7D)
Hence,
![\dfrac{A}{B}=\dfrac{4^2}{7^2}----(1)](https://tex.z-dn.net/?f=%5Cdfrac%7BA%7D%7BB%7D%3D%5Cdfrac%7B4%5E2%7D%7B7%5E2%7D----%281%29)
Also the sum of their areas is 65 cm^2.
i.e. ![A+B=65\\\\A=65-B----(2)](https://tex.z-dn.net/?f=A%2BB%3D65%5C%5C%5C%5CA%3D65-B----%282%29)
on putting equation (2) in equation (1) we have:
![\dfrac{(65-B)}{B}=\dfrac{4^2}{7^2}=\dfrac{16}{49}](https://tex.z-dn.net/?f=%5Cdfrac%7B%2865-B%29%7D%7BB%7D%3D%5Cdfrac%7B4%5E2%7D%7B7%5E2%7D%3D%5Cdfrac%7B16%7D%7B49%7D)
49(65-B)=16B
49×65-49B=16B
49×65=49B+16B
49×65=65B
B=49 cm^2
Hence by equation (2) we have:
A=65-B
A=65-49=16 cm^2.
Hence the area of each of the triangles are:
49 cm^2 and 16 cm^2.
Answer:
Check the explanation
Step-by-step explanation:
1. FALSE. If X and Y are 2 solutions to the equation AX = b, then A(X+Y) = AX +AY = b+b = 2b ≠ b as b ≠ 0. This implies that X+Y is not a solution to the equation AX = b. Hence the set of solutions to the equation AX = b is not closed under vector addition.
2. FALSE. If X is a solution to the equation AX = b and if k is an arbitrary scalar other than 1, then A(kX) = kAX = kb≠ b. This implies that kX is not a solution to the equation AX = b. Hence the set of solutions to the equation AX = b is not closed under scalar multiplication.
3. FALSE, unless b = 0. A.0 = 0 . Since b ≠ 0, the set of solutions to the equation AX = b does not contain the 0 vector.
4. FALSE. In view of 1,2,3 above, the set of solutions to the equation AX = b is not a subspace since b ≠ 0.