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professor190 [17]
3 years ago
9

Consider this system of two linear equations.

Mathematics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

Step-by-step explanation:

so these two lines cross at some point.. and they want to know the y point at which they cross... sooo..

solve them as a system to find the points that are common to both

Eq. 1)  5x -2y = 16

Eq. 2) -4x+6y=-15

i'll use the elimination method as these equations "lend" themselves to that way.. they are easy to solve that way .. in other words..

multiply Eq. 1 by 3  the add Eq. 2 .. to it... add the terms as columns....

    15x-6y = 48

<u> + ( -4x+6y=-15)</u>

=    11x+ 0y =33

x=3    

nice, now plug 3 in for x in one of the Eq.   try Eq. 1

5(3)-2y=16

15-2y=16

-2y =1

y = -\frac{1}{2}

since we're looking for the y coordinate...  that's your answer -\frac{1}{2}

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Thus, options A and D hold, from the simplifications above.

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\begin{gathered} \text{For option B)} \\ \text{substitute for n=1 into the expression n}^2+3n+2n,\text{ we have} \\ 1^2+3(1)+2(1)=1+3+2=6 \\ \text{substitute for n=1 into the expression 6n, we have} \\ 6(1)=6 \\ \text{Thus, the expression n}^2+3n+2n\text{ is equivalent to 6n, for n=1} \end{gathered}\begin{gathered} \text{For option C)} \\ \text{The expression n}^2+3n+2n\text{ does not simplify to 7n} \end{gathered}\begin{gathered} \text{For option E)} \\ \text{substitute for n=4 into the expression n}^2+3n+2n,\text{ we have:} \\ 4^2+3(4)+2(4)=16+12+8=36 \\ \text{substitute for n=6 into the expression 6n, we have:} \\ 6(4)=24 \\ \text{Thus, the two(2) expressions are not equivalent to each other, for n=4} \end{gathered}\begin{gathered} \text{For option F)} \\ \text{substitute for n=3 into the expression n}^2+3n+2n,\text{ we have:} \\ 3^2+3(3)+2(3)=9+9+6=24 \\ \text{substitute for n=3 into the expression 6n, we have:} \\ 6(3)=18 \\ \text{Thus, the two(2) expressions are not equivalent to each other, for n=3} \end{gathered}

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creativ13 [48]
Are you referring to this question?
During a dig, an archaeological team starts at an elevation of −5 1/2 feet. At a rate of 2 3/4 feet per hour, the team digs deeper into the surface for 3 1/2 hours. For the next 4 1/2 hours, the team digs at a rate of 1 5/12 feet per hour. Then the team quits for the day.
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<span>What was the team's elevation at the end of the day?
</span>
If so, then let us find out the elevation.

The rate of their dig is 2 3/4 ft per hr for 3 1/2 hours, then the total distance dug after 3 1/2 hours should be: 9 5/8 ft

Here’s how we get the total distance:

 First, change mixed number into improper fraction and proceed to multiplication

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Multiply.

11/4  x  7/2  = 77/8

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Let us move on at the rate of 1 5/12 ft per hr for the next 4 1/2 hr,

Following the same procedure as above, the distance would be: 6 9/24 ft

 

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Thus,to sum up everything, they have a total dug of:

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Then, their elevation on that day<span> is -5 1/2 - 16 = -21 1/2 ft</span>

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