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stellarik [79]
3 years ago
7

19−6(−k+4) Simplify to create an equivalent expression

Mathematics
2 answers:
almond37 [142]3 years ago
8 0

\text{Hey there!}\\\\\\\text{In order for you to solve this equation you need to distribute}\\\text{-6(-k) = 6k \& -6(4) = -24}\\\\\text{New equation: 19 + 6k + (-24)}\\\\\text{Combine your like terms (if there is any)}\\\text{(6k(19+(-24)))}\\\text{Your like terms are:  19  + (-24) = -5}

\text{Your new equation is: 6k + (-5)}\\\\\text{+ \&  -  =  -}\text{\underline{ (positive \& negative equals negative)}}\\\\\text{The number in the center will be a negative}\\\\\boxed{\boxed{\text{Answer: 6k - 5}}}\checkmark\\\\\\\\\\\text{Good luck on your assignment and enjoy your day!}\\\\\frak{LoveYourselfFirst:)}

JulijaS [17]3 years ago
7 0

Answer:

19+6k-24= 6k-5

Step-by-step explanation:

work is shown

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Answer:

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The level of a lake was 8 inches below normal. It decrease 1 1/4 inches in June and 2 3/8 inches more in July. What was the new
zloy xaker [14]

Answer:

Step-by-step explanation:

1

the initial level is below normal so it is represented by a negative number. the level continued to decrease in June and July so those changes are also represented by negative numbers. find the sum of these values to find what the new level was with respect to the normal level

2

write the mixed numbers as improper fractions

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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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Answer:

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Step-by-step explanation:

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7 0
3 years ago
Line ℓ1 has the equation
aleksley [76]

Answer:

\sqrt{2} units

Step-by-step explanation:

The Line 1 has equation x + y = 5 ....... (1) , and  

Line 2 has equation x + y = 3 .......... (2)

Now, we have to find the perpendicular distance between line 1 and line 2.

We can say that (3,0) is a point on line 2 as it satisfies the equation (2).

Now, the perpendicular distance from point (3,0) to the line 1 will be given by the formula

\frac{|3 + 0 - 5|}{\sqrt{1^{2}+ 1^{2}}} = \frac{2}{\sqrt{2}} = \sqrt{2} units (Answer)

We know that the perpendicular distance from any external point (x_{1},y_{1}) to a given straight line ax + by + c = 0 is given by the formula \frac{|ax_{1} + by_{1} + c|}{\sqrt{a^{2} + b^{2}}}.

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3 years ago
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