What will be the length of the other diagonal of a rhombus if one of the diagonals is 16 cm and the perimeter of the rhombus is 68cm?
Continued…Substituting the values17^2= 8^2 +x^2289=64+x^2289–64 = x^2225 = x^2+15 or -15 = x (taking sq.root)Discarding negative signLength of the other diagonal DB= 2x = 2*15= 30cmAns : 30 cm
What must be added to each term of the ratio of the 49:68, so that it becomes 3:4?
Let x be added to each term(49 + x) (4) = (3) (68+x)x = 8
What is the correct answer to this Mensa puzzle?
I think an answer of 112 might be suitable for this question. I reasoned as follows.The initial part is same as given in most answers here.Looking at the first 2 columns we have60/5=1268/4=1752/4=1381/9=9Here, we have a sequence (12,17,13,9)Looking at columns 4 and 3,we have48/4=1254/3=1878/6=13?/14=xThis gives an unfinished sequence (12,18,13,x)If we identify x, it is as good as finding the missing number. Clearly, it has to be divisible by 14 according to our established logic. But we can impose another constraint by looking at the sequence we obtained earlier (12,17,13,9). Here, we have a jump of 5 at first followed by decreases of 4 at a time. Now our incomplete sequence is (12,18,13,x). Here, we have a jump of 6 at first followed by a decrease of 5. Comparing with the pattern observed in the previous sequence, we can say that x=13-5=8.Now, the missing number= 14*8=112But since this is not there among the choices, it would be better to go with 84 (as in most answers) since it is the only one that satisfies the condition of divisibility by 14 at least.
How to factorize x^4-3x+20 Please provide steps?
x^4 − 3x + 20 First, I know it will break down into the product of two terms: (x² + bx + c) • (x² + dx + [20 ⁄ c]) Then I "guessed" that "c" is either 4 or 5 (it doesn't matter) (x² + bx + 4) • (x² + dx + 5) Then I multiplied it out and I know the "x²" term must be zero and the "x" term must be "-3x" I got [b = -3] and [d = 3] (x² − 3x + 4) • (x² + 3x + 5) Each of these two quadratic terms again factor. The roots however are imaginary. If you use the formula to solve for the roots of a quadratic equation you can easily find them.
I need like someone whos real good at math please!!?
okay so i have this challenge thing in my summer math packet thing and this is my last question(s) and i rly dont get it at all. 1. how many 3/10 (three tenths) centimeter segments are in 3 centimeters _______segments 2.how many 3/10 centimeter segments are in 4 1/5 ( four and one fifths) centimeters __________segments 3. how many 4/10 ( fourth tenths) centimeter segments are in 6 4/5 ( six adn four fifths) centimeters. _________segments thnks