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mars1129 [50]
3 years ago
6

Distinguish whether the ordered pair (16, -3) is a solution to:

Mathematics
1 answer:
sammy [17]3 years ago
5 0
I believe you should try to plug in the ordered pair to see if the solutions work.

The ordered pair is (x,y) because x is on the x-axis and y is on the y-axis, therefore in (16,-3) the x=16 and y=-3.

Now substitute, meaning plug in those numbers into the equation.
For x+2y=10 would be 16+2(-3)=10
Now you just need to solve one side and see if it equals the other side.
16+2(-3)
First you use order of operations to solve this. PEMDAS. So you multiply 2(-3) first because of the parenthesis and it being multiplication.
16+(-6) and when you have a positive number adding a negative number it’s going backwards of the number line, basically meaning subtraction in a way. Sorry if this confuses you, if you already know how to do negatives and such nevermind this part.
But 16+(-6)=10
So now you look at both side of the equation, does the left side equal to the right? 10=10, so yes. It is a solution for that equation.

Now for the next equation, 7y=-21
Again, plug in the ordered pair (16,-3) into the equation. Remember that it’s (x,y).
There is no x in this equation so no need to worry about that; you only plug in y for this one.
7(-3) Now you multiply. Whenever you multiply a positive number and a negative number, the answer will always be negative. So 7(-3) is -21.
Now look if the left side is equivalent to the right. Does -21=-21? Yes. The ordered pair is a solution.

(16,-3) is a solution to both equations.

Hope this helps!
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Axis of sym: x =
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Answer:

<h2>SEE BELOW</h2>

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • quadratic function
  • PEMDAS
<h3>let's solve:</h3>

vertex:(h,k)

therefore

vertex:(-1,4)

axis of symmetry:x=h

therefore

axis of symmetry:x=-1

  • to find the quadratic equation we need to figure out the vertex form of quadratic equation and then simply it to standard form i.e ax²+bx+c=0

vertex form of quadratic equation:

  • y=a(x-h)²+k

therefore

  • y=a(x-(-1))²+4
  • y=a(x+1)²+4

it's to notice that we don't know what a is

therefore we have to figure it out

the graph crosses y-asix at (0,3) coordinates

so,

3=a(0+1)²+4

simplify parentheses:

3 = a(1 {)}^{2}  + 4

simplify exponent:

3 =  a + 4

therefore

a =  - 1

our vertex form of quadratic equation is

  • y=-(x+1)²+4

let's simplify it to standard form

simplify square:

y =  - ( {x}^{2}  + 2x + 1)  + 4

simplify parentheses:

y =  -  {x}^{2}  - 2x - 1 + 4

simplify addition:

y =  -  {x}^{2}  - 2x + 3

therefore our answer is D)y=-x²-2x+3

the domain of the function

x\in \mathbb{R}

and the range of the function is

y\leqslant 4

zeroes of the function:

-  {x}^{2}  - 2x + 3 = 0

\sf divide \: both \: sides \: by \:  - 1

{x}^{2}  + 2x - 3 = 0

\implies \:  {x}^{2} +   3x  - x  +  3 = 0

factor out x and -1 respectively:

\sf \implies \: x(x + 3)   - 1(x  + 3 )= 0

group:

\implies \: (x - 1)(x + 3) = 0

therefore

\begin{cases} x_{1} = 1 \\  x_{2}  =  - 3\end{cases}

4 0
3 years ago
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