factor the polynomial. 17xy^4 51x^2y^3 a) 7xy (10y^4 44xy^2) b) 17xy^3 (y 3x) c) 17xy (1 1/3xy^2) d) xy (17 51xy^2)

PLEASE HELP I AM ON A TIME LIMIT

Alinara [238K]

Answer:

(a) even: J, K, M, O; odd: L, N

(b) L and O are connected to J

(c) N is of degree 3

Step-by-step explanation:

Count the edge ends that intersect each vertex. You get ...

J-2, K-2, L-3, M-2, N-3, O-4

These numbers are the <em>degree</em> of the vertex.

a) Vertices with even degree are J, K, M, O, since these have degrees of 2, 2, 2, and 4, respectively--all even numbers.

Vertices with odd degree are L and N, since these both have degree 3, an odd number.

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b) The vertices that are adjacent to J are the ones at the other ends of the edges that intersect vertex J. One of those two edges connects to vertex L; the other to vertex O. J is adjacent to L and O.

__

c) When we counted edges in part (a), we found vertex N to be of degree 3.

5 0

3 years ago

Peter ran 15 miles without stopping during the marathon. Anne ran the distance that Peter ran before she stopped. How far did

Anika [276]

The Answer Should be C

8 0

4 years ago

ASAP PLZ HELP 1 EASY QUESTION HELP :/

MrMuchimi

If 2 triangles are similar, the ratio of one side of the triangle to the corresponding side of the other is always same for all 3 sides.
We can make use of this rule to calculate the length of sides.
From the image,
3 / 15 = 6 / x
Simplify,
1 / 5 = 6 / x
Use cross method to solve this.
5 x 6 = 1 x x
X = 30
Therefore the length of x is 30ft., A.

6 0

3 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$