<h2>hope it helps you please mark me as brainliest </h2>
Answer:
There are loads, but for example you can find the diagonal from a lamppost the the end of its shadow if you know the height of the post and the length of the shadow.
Hope I helped x
Answer:
domain: {x | x is a real number}
range: {y l y> -8}
Step-by-step explanation:
f(x) = 4x² – 8 is a parabola, a U shape.
Since the stretch factor, 4, is positive, it opens up, there it will have a minimum value, the lowest point in the parabola.
y > -8 because the minimum is -8.
Parabolas do not have restricted "x" values. "4" does not restrict x because it is the stretch factor, which determines how wide the parabola is.
Quadratic standard form:
f(x) = ax² + bx + c
"a" represents how wide the graph is. If it's negative it opens down, if it's positive it opens up.
"b", if written, tells you it is not centred on the y-axis. It is not written, so the vertex is on the y-axis.
"c" is the y-intercept. In this case, since b = 0, it is also the minimum value.
Answer:
for the first one, simply add g(x) and h(x) :
x+3 + 4x+1 = 5x + 4
the second one, you would multiply them :
(x+3)(4x+1) = 4x^2 + 13x + 3
the last one, you would subtract :
(x+3)-(4x+1) = -3x + 2
and then substitute 2 for 'x' :
-3*2 + 2 = -6 + 2 = -4
