Answer:
A
Step-by-step explanation:
This option works
The <em>additional information</em> needed to prove that both triangles are congruent by the SSS Congruence Theorem would be: <em>C. HJ ≅ LN</em>
<em>Recall:</em>
- Based on the Side-Side-Side Congruence Theorem, (SSS), two triangles can be said to be congruent to each other if they have three pairs of congruent sides.
Thus, in the two triangles given, the two triangles has:
- Two pairs of congruent sides - HI ≅ ML and IJ ≅ MN
Therefore, an <em>additional information</em> needed to prove that both triangles are congruent by the SSS Congruence Theorem would be: <em>C. HJ ≅ LN</em>
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Learn more about SSS Congruence Theorem on:
brainly.com/question/4280133
Its in the hundreds column
<h3>
Answer: choice C) 15</h3>
Simplify the left side to get
2(4+x)+(13+x)
2(4)+2(x) +13+x
8+2x+13+x
3x+21
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So the original equation
2(4+x)+(13+x) = 3x+k
turns into
3x+21 = 3x+k
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Subtract 3x from both sides
3x+21 = 3x+k
3x+21-3x = 3x+k-3x
21 = k
k = 21
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If k = 21, then the original equation will have infinitely many solutions. This is because we will end up with 3x+21 on both sides, leading to 0 = 0 after getting everything to one side. This is a true equation no matter what x happens to be.
If k is some fixed number other than 21, then there will be no solutions. This equation is inconsistent (one side says one thing, the other side says something different). If k = 15, then
3x+21 = 3x+k
3x+21 = 3x+15
21 = 15 .... subtract 3x from both sides
The last equation is false, so there are no solutions here.
note: if you replace k with a variable term, then there will be exactly one solution.
The given equation is of the form y = m x + b where m is the slope and b is the y intercept that is value of y when x =0 , or we can say b is the initial value.
In the given equation which is y = 4t+2, on comparing we will get m =4 that is slope is 4 and here slope represents the rate or speed and 2 is the y intercept that is b. And 2 represents the initial position that is, the ant is 2 units from the home initially .
