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lbvjy [14]
2 years ago
9

Consider the equation 4x2 + 2x = (x + 2)(x − 5)x Match the Property used to create an equation with the same solution set. optio

ns: Commutative Property, Multiplication Property,associative Property, Division Property,Distributive Property What property matches up with the equations?
1. (4x2 + 2x) = (x + 2)((x − 5)x)
2. 2x + 4x2 = x(x − 5)(x + 2)
3. x(4x + 2) = (x + 2)(x2 − 5x)
4. 12x2 + 6x = (x + 2)(x − 5)3x
5. 4x + 2 = (x + 2)(x − 5)
Mathematics
1 answer:
mel-nik [20]2 years ago
8 0

Factor then solve. Exact Form: x = 0 , 7 + √ 97 2 , 7 − √ 97 2 Decimal Form: x = 0 , 8.42442890 … , − 1.42442890 …

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Answer:

We show 1 to distinguish a vertex, but with laws of supplementary the answer is 2.

Step-by-step explanation:

7 0
2 years ago
A swimming pool is represented on a coordinate grid with the following vertices: Vertex A is at (−2,−3) . Vertex B is at ​​ (−2,
Semmy [17]

Answer:

26 yards.

Step-by-step explanation:

We can find the distance walked by Jared by finding the distances between each of the given vertex of swimming pool. We will use distance formula to find the distances between each vertex.  

\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

\text{Distance between vertex A and vertex B}=\sqrt{(-2--2)^{2}+(-3-5)^{2}}  

\text{Distance between vertex A and vertex B}=\sqrt{(-2+2)^{2}+(-8)^{2}}

\text{Distance between vertex A and vertex B}=\sqrt{(0)^{2}+(-8)^{2}}

\text{Distance between vertex A and vertex B}=\sqrt{64}=8

Now let us find distance between vertex B and vertex C.

\text{Distance between vertex B and vertex C}=\sqrt{(-2-3)^{2}+(5-5)^{2}}

\text{Distance between vertex B and vertex C}=\sqrt{(-5)^{2}+(0)^{2}}

\text{Distance between vertex B and vertex C}=\sqrt{25}=5

Now let us find distance between vertex C and vertex D.

\text{Distance between vertex C and vertex D}=\sqrt{(3-3)^{2}+(5--3)^{2}}

\text{Distance between vertex C and vertex D}=\sqrt{(0)^{2}+(5+3)^{2}}

\text{Distance between vertex C and vertex D}=\sqrt{64}=8

Now let us find distance between vertex D and vertex A.

\text{Distance between vertex D and vertex A}=\sqrt{(3--2)^{2}+(-3--3)^{2}}  

\text{Distance between vertex D and vertex A}=\sqrt{(3+2)^{2}+(-3+3)^{2}}    

\text{Distance between vertex D and vertex A}=\sqrt{(5)^{2}+(0)^{2}}

\text{Distance between vertex D and vertex A}=\sqrt{25}=5

Now let us add all these distances to find the total distance walked by Jared.  

\text{Total distance walked by Jared}=8+5+8+5

\text{Total distance walked by Jared}=26  

Since we have been given that one unit on the grid equals 1 yard,   therefore, Jared walked a distance equal to 26 yards.  

3 0
3 years ago
A construction crew is lengthening a road that originally measured 47 miles. The crew is adding one mile to the road each day. L
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Answer:

78 miles

Step-by-step explanation:

Given that:

Original length, L = 47 miles

Additional length (miles) added per day, = 1 mile

Representing as an equation :

L(D) = original length + additional length per day * number of days

Let, D = number of days

L(D) = 47 + D

Length after 31 days :

L(31) = 47 + 31

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In numbers only: 
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Find 100/20 x 30
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