Answer:
JK=7
Step-by-step explanation:
From the line segment, since J is on it ,it means the line segment is
represented as I J K
from this illustration, we can say that the longest part of the line segment is from I to K
this means that, IJ +JK =IK
making JK the subject,
JK= IK - IJ
but from the question, JK=2x-1 , IK=3x+2 and IJ=3x-5
substituting them in the expression,
2x-1 =3x+2 -(3x-5)
solving for x
2x-1 =3x+2-3x+5
2x-1 =0+7
2x-1 =7
2x=1+7
2x=8
dividing through by 2
2x/2 =8/2
x=4
but the question says we should find the numerical value for JK
but from the line segment,
JK=2x-1
but now we know the value of x to be 2
so substituting it in the formula
JK= 2(4)-1
JK=8-1
JK=7
therefore, the numerical value for JK is 7
Answer:
He multiplyed 3x8 and 5x8 when it should be 3x8 and 5x1 because you but 1 under the 8 to make it into a fraction.
Step-by-step explanation:
Answer:
B. False, they could be vertical angles
Step-by-step explanation:
Supplementary just mean when two angles form 180° (a straight line).
So ∠F and ∠G must form a straight line, and ∠G and ∠H must form a a straight line.
I drew the image out below and as you can see, angle F and H forms a vertical angle. A vertical angles is when two lines cross, the opposite angles from each others are called vertical angles.
Standard form is ax+by=c
Steps:
y + 4 = 1.5(x - 4)
y + 4 = 1.5x - 6
-1.5x + y = -10 is the standard form.
Answer:
see attached
Step-by-step explanation:
The grid show the non-possibilities in red, with each number corresponding to the statement that eliminates that choice. The green square (with black text) shows the one combination that is specified already (by statement 4). The lighter green numbers show possible alternatives: first period may be Schiller or English, and room 113 will be the other one. Similarly, Art may be 3rd period or Thomlinson.
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These choices (light green 5, light green 6) give rise to four possibilities. Working through them, you run into inconsistencies if you choose Schiller for first period. (Art must be, but can't be, in room 112.) That leaves two possibilities.
Again, you run into inconsistencies if you choose Thomlinson as the Art teacher. (The class in 112 is 2 periods after Xavier's class, not 1.)
Hence, the only viable pair of remaining choices is Schiller in room 113 and art in 3rd period.
The final schedule is shown in the attachment.
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<em>Additional comment</em>
When I'm working these on paper, I use an X to mark any impossible combinations, and a circle (O) to mark a known combination. In any given 4×4 square of the grid, the remaining cells of the row and column containing a O must be Xs. Consistency must be maintained between rows and columns. This often means filling a circle in one place may result in a circle being filled in another place. Of course, once 3 of the squares in a row or column have Xs, the remaining one must be O.