If There is a math equation you can use math-way.
Also it would help if you can take a better picture.
Once you take a better picture I promise i will try my best to answer the question.
Answer:
<h2>SEE BELOW</h2>
Step-by-step explanation:
<h3>to understand this</h3><h3>you need to know about:</h3>
<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:
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:

simplify exponent:

therefore

our vertex form of quadratic equation is
let's simplify it to standard form
simplify square:

simplify parentheses:

simplify addition:

therefore our answer is D)y=-x²-2x+3
the domain of the function

and the range of the function is

zeroes of the function:




factor out x and -1 respectively:

group:

therefore

Answer:
Los lados del triángulo rectángulo miden 3, 4 y 5, respectivamente.
Step-by-step explanation:
Un triángulo rectángulo puede ser descrito mediante el teorema de Pitágoras, para el caso de tres lados representando tres números enteros consecutivos, tenemos que:
(1)
Donde
es un número natural.
A continuación, expandimos la expresión y resolvemos:


La única solución factible es
. En consecuencia, los lados del triángulo rectángulo miden 3, 4 y 5, respectivamente.
Recall that 2sin(x) cos(x) is actually equal to sin(2x).
We can prove this by expanding sin(2x) to sin(x + x).
sin(x + x) = sin(x) cos(x) + cos(x) sin(x) = 2sinxcosx
Thus, 2sin(x/2)cos(x/2) can be rewritten in the form:
sin(2x/2), and this simplifies down to sinx.
Answer:
66 and 62
Step-by-step explanation:
128 divided by 2 gives you 64, and by taking two away from one number and giving it to the other, you're final answers are 66 and 62