Answer:
Step-by-step explanation:
Use BODMAS rule :
Bracket , Of , Division , Multiplication , Addition , Subtraction
![\sf{2 \times [ 34 - 5 \times (3 + 1) ] }](https://tex.z-dn.net/?f=%20%5Csf%7B2%20%5Ctimes%20%5B%2034%20-%205%20%5Ctimes%20%283%20%2B%201%29%20%5D%20%7D)
Add the numbers : 3 and 1
![\dashrightarrow{ \sf{2 \times [ 34 - 5 \times 4 ] }}](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%7B%20%5Csf%7B2%20%5Ctimes%20%5B%2034%20-%205%20%5Ctimes%204%20%5D%20%7D%7D)
Multiply the numbers : 5 and 4
![\dashrightarrow{ \sf{2 \times [ 34 - 20 \: ] }}](https://tex.z-dn.net/?f=%20%5Cdashrightarrow%7B%20%5Csf%7B2%20%5Ctimes%20%5B%2034%20%20-%2020%20%5C%3A%20%5D%20%20%7D%7D)
Subtract 20 from 34
Multiply the numbers : 2 and 14
Hope I helped!
Best regards! :D
Answer:
2
Step-by-step explanation:
x^0+ y^0
Let x = 3 and y = 2
3^0 + 2^0
Raised to the 0 power is 1
1 + 1
2
A. 
To find greater than or smaller than relation, we multiply the terms like (numerator of L.H.S with denominator of R.H.S and put the value on the left side. Then multiply the denominator of L.H.S with numerator of R.H.S and put the value on right side. Now compare the digits.)
So, solving A, we get 810<209 ... This is false
B. 
= 238>589 ..... This is false
C. 
= 496>780 .... This is false
D. 
= 420<660 ..... This is true
Hence, option D is true.
Answer:
<em>The second figure ( rectangle ) has a longer length of it's diagonal comparative to the first figure ( square )</em>
Step-by-step explanation:
We can't confirm the length of these diagonals based on the appearance of the figure, so let us apply Pythagorean Theorem;
This diagonal divides each figure ( square + rectangle ) into two congruent, right angle triangles ⇒ from which we may apply Pythagorean Theorem, where the diagonal acts as the hypotenuse;
5^2 + 5^2 = x^2 ⇒ x is the length of the diagonal,
25 + 25 = x^2,
x^2 = 50,
x = √50
Now the same procedure can be applied to this other quadrilateral;
3^2 + 7^2 = x^2 ⇒ x is the length of the diagonal,
9 + 49 = x^2,
x^2 = 58,
x = √58
<em>Therefore the second figure ( rectangle ) has a longer length of it's diagonal comparative to the first figure ( square )</em>
