Answer:
q≤-26
Step-by-step explanation:
7.85+q≤-18.15
-7.85 -7.85
q≤-26
It would be: -7/12 + 7/10 = -35+42 /60 = 7/60
So, your final answer is 7/60
Hope this helps!
Answer:
see below
Step-by-step explanation:
For simplifying expressions of this sort, there are four rules of exponents that come into play;

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I find it convenient to eliminate the fractions by adding the exponents, then rewrite any negative exponents as denominator factors.

Answer:
y=1x+1
Step-by-step explanation:
y=1x+1 gradient= 1 and y-intersept = 1
Answer:
The correct answer is 24
Step-by-step explanation:
to solve this you will need to use the pathagreom theorum
a^{2}+b^{2}=c^{2}
A= one side lenth
B= the secons side lenth
C= hypotnuse
It is helpfull to draw out the situation
you know that the latter is 25 ft, that is your hypotnuse
you also know that the 7 ft away from the base of the building is one of the side lenths, lets call it side a
so plug the numbers into the equation
7^2 + b^2 = 25 ^2
you leave b^2 alone because that is the side you are trying to find
now square 7 and 25 but leave b^2 alone
49 + b^2 = 625
now subtract 49 from both sides
b^2 = 576
now to get rid of the square of b you have to do the opposite and square root both sides removing the square of the B and giving you the answer of..........
B= 24
Hope this helped!! I tryed to explain it as simpil as possiable