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Vladimir79 [104]
2 years ago
8

Solving a System of Linear Equations: Substitution

Mathematics
1 answer:
CaHeK987 [17]2 years ago
3 0

Answer:

where is the picture of the tables

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What is the value of 6^0
Tema [17]

Answer:

1

Step-by-step explanation:

Any base with an exponent of 0 equals to 1.

6^0 = 1

7 0
3 years ago
Read 2 more answers
How do you solve this problem? (x+3) cubed
Lubov Fominskaja [6]
You would expand becausee it isn't equal to anything

(x+3)^3=(x+3)(x+3)(x+3)
we can use binomial theorem or pascal's triangle
anyway, pascal's triangle is more practicl for this one

so
for the 3rd power
(a+b)^3=1a^3+3a^2b^1+3a^1b^2+1b^3

a=x
b=3

(x+3)^3=1x^3+3x^2(3^1)+3x^1(3)^2+1(3)^3=
x^3+9x^2+27x+27


7 0
3 years ago
Read 2 more answers
PLEASE HELP I WILL GIVE EXTRA POINTS
Evgen [1.6K]

Answer:

  C

Step-by-step explanation:

A relation is a function if each input gives exactly one output. There is one price for each number of pounds, so the relation is a function.

__

<em>Additional comment</em>

Relations expressed in terms of ordered pairs or tables can be identified as "not a function" if any input (x) value is repeated. Relations expressed as a graph will be "not a function" if any vertical line intersects the graph at more than one point. (This is called "the vertical line test.")

The cost function in this case is a straight line through the origin. It has a slope of $5.99/lb. Any polynomial relation will be a function.

6 0
2 years ago
What are the number of integral solutions of the equation 2x+2y+z=20 such that x&gt;=0 , y&gt;=0 , z&gt;=0?
zaharov [31]
2x+2y+z=20\\&#10;z=\dfrac{20}{2x+2y}\\&#10;z=\dfrac{10}{x+y}

Now, for z to be an integer, the sum x+y must be a divisor of 10. 
It has to be a positive divisor since z≥0. Also x≠y.

x+y=1 \\&#10;x=1-y\\&#10;
x≥0 and y≥0 so y can be equal to either 0 or 1. There are 2 solution in this case.

x+y=2\\&#10;x=2-y\\
In this case, y can be equal to 0,1, or 2, but for y=1 ⇒ x=1, so there are two solutions.

x+y=5\\&#10;x=5-y
y can be 0,1,2,3,4 or 5 - 6 solutions

x+y=10\\&#10;x=10-y
y can be 0,1,2,3,4,5,6,7,8,9,10, but for y=5 ⇒ x=5, so 10 solutions.

2+2+6+10=20 solutions in total.
6 0
3 years ago
Question 2 (Multiple Choice Worth 5 points)
steposvetlana [31]
Number of remaining rose cards = 6
number of remaining cards = 8
P(rose card) = 6/8

P(blue marble) = 18/50
P(tail) = 30/50
P(blue marble and tail) = 18/50 x 30/50 = 540/2500
6 0
3 years ago
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