Answer:
JK = 30 .
Step-by-step explanation:
As JMLK is a kite the diagonals bisect each other at right angles so triangle JNK is a right triangle.
JK^2 = 18^2 + 24^2 = 900
JK = 30.
Answer:
40 daps
Step-by-step explanation:
lets say daps is d, yaps are y and baps are b
4d = 7y, 5y=3b
First convert to y
4/7d = y
Now convert that to b
5(4d/7) = 3b
20d / 7 = 3b
20d / 21 = b
Plug it in:
42(20d/21)
40d
aka 40 daps
Answer:
75% of the data will reside in the range 23000 to 28400.
Step-by-step explanation :
To find the range of values :
We need to find the values that deviate from the mean. Since we want at least 75% of the data to reside between the range therefore we have,
Solving this, we would get k = 2 which shows the value one needs to find lies outside the range.
Range is given by : mean +/- (z score) × (value of a standard deviation)
⇒ Range : 25700 +/- 2 × 1350
⇒ Range : (25700 - 2700) to (25700 + 2700)
Hence, 75% of the data will reside in the range 23000 to 28400.
Answer: No solution
Step-by-step explanation:
9-(-8)=20 given
9+8=20 the two negatives cancel out, making the eight positive
17=20 this statement is not correct, so the answer is no solution
no solution
I'm not sure if there was supposed to be a variable in there somewhere, but I did my best to explain. Hope this helps!
9514 1404 393
Answer:
C, A, A
Step-by-step explanation:
In general, you ...
- identify the coefficients of one of the variables
- swap them, and negate one of them
- multiply the corresponding equations by the "adjusted" coefficients.
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In problem 1, the x-coefficients are 8 and 2. A common factor of 2 can be removed so that we're dealing with the numbers 4 and 1. Assuming we want to multiply one of the equations by 1, leaving it unchanged, the value we want to multiply by will be -4. After we swap the coefficients, that multiplier is associated with equation 2:
multiply equation 2 by -4 . . . (eliminates x)
Likewise, the y-coefficients in problem 1 are -1 and 3. Again, if we want to multiply one of the equations by 1, leaving it unchanged, the coefficient we will change the sign of is -1 (becomes 1). After we swap the coefficients, the multiplier 3 is associated with equation 1:
multiply equation 1 by 3 . . . (eliminates y)
These two choices are B and A, respectively, so the one that does NOT work for problem 1 is choice C, as indicated below.
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The other problems are worked in a similar fashion.