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Minchanka [31]
3 years ago
14

How many times can 36 go into 79

Mathematics
1 answer:
Stella [2.4K]3 years ago
7 0
2.2 is the answer because when you divide its a decimal
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Slope of (-2,2) and (3,4)
liq [111]

Answer:

2/5

Step-by-step explanation:

Good luck!

7 0
3 years ago
What two numbers add to -36 and<br> multiply to 35?
andreyandreev [35.5K]
-1, -35.

When you multiply two negative numbers, you get a positive result.
3 0
3 years ago
Read 2 more answers
Which point fits the description: the y- coordinate is half the x-coordinate?
Naily [24]

Answer:

3,6

Step-by-step explanation:

8 0
3 years ago
What is the solution of x=2+ square root x-2?
alex41 [277]

So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

x-2=\sqrt{x-2}

Next, square both sides:

(x-2)^2=x-2\\(x-2)(x-2)=x-2\\x^2-4x+4=x-2

Next, subtract x and add 2 to both sides of the equation:

x^2-5x+6=0

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

x^2-2x-3x+6=0

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

x(x-2)-3(x-2)=0

Now you can rewrite the equation as (x-3)(x-2)=0

Now, apply the Zero Product Property and solve for x as such:

x-3=0\\x=3\\\\x-2=0\\x=2

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

2=2+\sqrt{2-2}\\2=2+\sqrt{0}\\2=2+0\\2=2\ \textsf{true}\\\\3=2+\sqrt{3-2}\\3=2+\sqrt{1}\\3=2+1\\3=3\ \textsf{true}

Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>

8 0
3 years ago
What is x+y=10;y=x+10
Alenkasestr [34]

[ Answer ]

\boxed{\bold{X \ = \ 0, \ Y \ = \ 10}}

[ Explanation ]

  • System Of Equations \bold{\left \{ {X \ + \ Y \ = \ 10} \atop {Y \ + \ X \ = \ 10}} \right.}

-----------------------------------

  • [Substitute] Y = X + 10

\bold{\begin{bmatrix}x+x+10=10\end{bmatrix}}

  • Isolate x for x + x + 10 = 10: x = 0

For y = 10

Substitute x = 0

Y = 0 + 10

0 + 10 = 10

y = 10

  • Solutions

X = 0

Y = 10

\boxed{\bold{[] \ Eclipsed \ []}}

3 0
3 years ago
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