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Liula [17]
2 years ago
9

3(2k - 5) = 6(k - 4) + 9

Mathematics
1 answer:
natita [175]2 years ago
6 0

Answer:

This expression has no numeric as the variable k gets factored out along with all of the numbers.

Step-by-step explanation:

3(2k - 5) = 6(k - 4) + 9

6k - 15 = 6k - 24 + 9

6k - 6k = 15 - 24 + 9\\

0 = 0

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Need help as soon as possible!<br><br> What is the value of g(3)?<br><br> 2<br> 3<br> 9<br> 14
Alex777 [14]

g(x) is a piecewise function in such a way that it changes how it's defined based on what x happens to be. There are three cases

Case A: g(x) = x-1 but only if -2 \le x < -1 (x is between -2 and -1; including -2 but excluding -1)

Case B: g(x) = 2x+3 but only when -1 \le x < 3 (x is between -1 and 3; including -1 but excluding 3)

Case C: g(x) = 6-x but only when x \ge 3

The input is x = 3 since we want to find the value of g(3). So we look at the 3 cases above (A,B,C) and determine that we use case C. Why? Because x = 3 makes x \ge 3 true. Put another way, x = 3 is in the interval [3, infinty). So we'll use g(x) = 6-x to find that...

g(x) = 6-x

g(3) = 6-3

g(3) = 3

Answer: 3

5 0
3 years ago
Can someone give me the answers asap! 15 points
Eduardwww [97]
In the first question,
Since D is the midpoint of line segment AC.
therefore, AD=DC.

In triangles, ABD & CBD.
Since, AB is congruent to CB and BD is a common side between those two triangles.
Therefore, AD is congruent to DC.

Finally, in triangles DEA & DEC
Since, ED line segment is a common side between those triangles and the measure of angle EDA is as the same of angle EDC which are equal to 90 degrees and AD is congruent to DC.
Therefore, triangle EDA is congruent to triangle EDC ( you have to take care of the triangle order letters in congruence) and EC is congruent to EA.
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3 years ago
Evaluate. Answer has to be in scientific notation<br><br> (8 x 10^6) + (7.3 x 10^4)
Ne4ueva [31]

Answer:

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What are the coordinates of the endpoints of the midsegment for △DEF that is parallel to DE¯¯¯¯¯?
Readme [11.4K]

Answer:

The endpoints of the midsegment for △DEF that is parallel to DE, are (-1,3.5) and (-1,2).

Step-by-step explanation:

If a line connecting the midpoint of two sides and parallel to the third side of the triangle, then it is called a midsegment.

From the given figure it is noticed that the vertices of the triangle are D(1,4), E(1,1) and F(-3,3).

If the midsegment is parallel to DE, then the end points of the midsegment are mid point of DF and EF.

Midpoint formula.

Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Midpoint of DF,

Midpoint=(\frac{1-3}{2},\frac{4+3}{2})

Midpoint=(-1,3.5)

Midpoint of EF,

Midpoint=(\frac{1-3}{2},\frac{1+3}{2})

Midpoint=(-1,2)

Therefore the endpoints of the midsegment for △DEF that is parallel to DE, are (-1,3.5) and (-1,2).

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3 years ago
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