Excercise 1:
No, this is not simplified fully. The full answer would be <span>−7dk+14d−21k</span>−7, not <span>14d - 9 - 21k - 7dk + 2. So, it is not equivalent.
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Excercise 4:
</span><span>7dk (2 - 3 - 1) - 7
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Just multiply 7dk into (2) (-3) and (-1)
you'd get:
<span><span>−<span>14dk</span></span>+</span>−<span>7 after simplifying it fully.
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p = parameter
w = width
w+2.75
p = 2w+2(w+2.75) = <span>4w</span>+<span>5.5 = 30
Answer to the last one:
w = 6.125</span>
Answer:
5 hours
Step-by-step explanation:
This is a classic pre-algebra question.

Answer:
<h2>x = 0, y = 5, z = 3</h2>
Step-by-step explanation:



Answer:
YET
Step-by-step explanation:
TO DECRYPT DKA USING THE CAESAR SHIFT CIPHER STARTING WITH 5 SHIFTS.
TO DO THIS DECRYPTION, WE'RE GOING TO LOOP THE ALPHABETS.
D = LOOPING THIS LETTER 5 TIMES, WE HAVE C - B - A - Z - Y.
SO, D IS DECRYPTED AS Y
GOING FURTHER, THE ENCRYPTION SHIFTS BY AN ADDITIONAL SPACE. THIS MEANS, IF THE FIRST "Y" WAS ENCRYPTED IN 5 SPACES TO GET D, AND AN ADDITIONAL SPACE WAS ADDED TO THE NEXT ALPHABET, IT BECOMES 6.
K IS THEN DECRYPTED AS
K = J - I - H - G - F - E
K IS DECRPYTED AS E
GOING FURTHER, THE NEXT WAS ENCRYPTED WITH AN ADDITIONAL SPACE EVEN MORE, SO, A IS DECRPYTED AS
A = Z - Y - X - W - V - U - T.
THUS, DKA IS DECRYPTED AS YET