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SCORPION-xisa [38]
3 years ago
13

Below are two parallel lines with a third line intersecting them.

Mathematics
1 answer:
Delicious77 [7]3 years ago
8 0

Answer:

x=50

Step-by-step explanation:

That is becasue they are in the same spot on different parallel lines.

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I will give the brainiest
Alisiya [41]

Answer:

A;C;E

Step-by-step explanation:

Let cost of almond per pound be X

Cost of 3 pounds of almonds=3X

cost of pistachios per pound = X+4

Cost of 2 pounds of pistachios=2X+8

Total cost=5X+8=48

5X+8=48

5X=40

X=8

cost of one pound almond is $8

cost of one pound pistachios is $12

Please mark it brainly

5 0
3 years ago
Vertex A in quadrilateral ABCD lies at (-3, 2). If you rotate ABCD 180° clockwise about the origin, what will be the coordinates
Marysya12 [62]
I think it might be B hope I’m right
3 0
4 years ago
Read 2 more answers
Algebraically determine if the relation x = y2 – 100 is symmetrical with respect to the x-axis, y-axis, or the origin.
yKpoI14uk [10]

Answer: Option D: x-axis symmetry

Step-by-step explanation:

The symmetry concept can be explained as below:

  1. If any point say, (x,y) on the graph, then (x,-y) is also on the graph, it shows x- axis symmetry.
  2. If any point say, (x,y) lies on the graph, then, (-x,y) should also lie on the same graph, then is shows y-axis symmetry.  

Now, here the graph is for x = y^{2}-100  is attached.

Suppose, (x,y) = (-100,0)

then, solving this point satisfies the equation of given graph.

Also, (x,-y) = (-100,0) satisfies the equation of the given graph.

Hence, the relation x = y^2 – 100 is symmetrical to x-axis.

4 0
4 years ago
On February 4th, Jamie and Kevin jogged together.
ludmilkaskok [199]

Find the common multiples of 3 and 4:

3: 3, 6, 9 12, 15

4: 4, 8, 12, 16

The smallest common multiple for both 3 and 4 is 12, so this means every 12 days they would run together.

Add 12 days to February 4th to get February 16th.

7 0
4 years ago
the three counters shown in the table are used for a board game. If the counters are tossed , how many ways can at least one cou
Elena L [17]

Total ways = {AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB}

Ways at least one counter with side A = {AAA, AAB, ABA, BAA, ABB, BAB, BBA}

Hence, the number of ways at least one counter with side A = 7.

8 0
3 years ago
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