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Lady bird [3.3K]
3 years ago
11

how do you determine if the answer to a problem is negative or positive in division multiplication addition and subtraction

Mathematics
2 answers:
erica [24]3 years ago
7 0
In multiplication, a negative times a positive has a negative answer. if both are positive or both are negative, the answer is positive.
5 * -5 = -25
-5* -5 = 25


in divisions say you have
-5/5 = -1
-5/-5= 1
5/-5= -1


in addition and subtraction, subtracting a negative means to add the two numbers. adding a negative means to subtract the second number.

5+ (-5) = 0
5- (-5) = 10
liq [111]3 years ago
3 0
I think the rules the same: n+n=n p+p=p and if its a n+p, the bigger one gets the sign. for example, if its -3+5, it's 2. basically, the negative 3 is another word for subtracting 3 from the 5. so, 3-5=2. another example is if its 3+(-5). this is the trick i made up to do these problems: so, 3-3=0. how much is left from the 5? 2. but its negative, so -2. this took like, 5 minutes to write, so...hope i helped! ;)
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After distributing, which equation is equivalent to -4(3x + 2) = 10a.-12x -8 = 10b.-12x + 8 = 10c.-12x -6 = 10d.-12x + 6 = 1
rodikova [14]

Answer:

Option A: -12x-8=10 is the correct answer.

Step-by-step explanation:

Given equation is:

-4 (3x+2) = 10

We will distribute the equation by multiplying the terms inside the brackets with -4.

Therefore,

-4*3x + 2*-4 = 10

-12x +(-8) = 10

-12x - 8 = 10

The equation -4 (3x+2) = 10 will become -12x-8=10 after distributing.

Hence,

Option A: -12x-8=10 is the correct answer.

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3 years ago
Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located
arsen [322]
Check the picture below.

\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

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