It'd be a lot easier to help you if you'd share the answer graphs.
Just so that you know:
<span>f(x) = log(x + 3) + 1 can be graphed as follows:
</span>
1. Graph y = log x. This function's domain is x>0. It begins in Quadrant IV and ends in Quadrant I. It intercepts the x-axis at (1,0).
2. Now move the entire graph 1 unit to the left. This gives you the graph of y = log (x+1).
3. Now translate this entire graph up 3 units. This gives you the graph of <span>f(x) = log(x + 3) + 1.</span>
what's the rest of the question am i graphing it or am i writing it in slope-intercept
Answer:
code = 56
Step-by-step explanation:
E = GCF(22,30) = 2
F = LCM(3,8) = 24
Given
3 * (7+10) = G +30 => 51 = G + 30 => G = 21
5 * (3+H) = 15 + 45 => 3+H = (60/5) = 12 => H = 12 - 3 = 9
Sum E + F + G + H = 2 + 24 + 21 + 9 = 56
Answer:
True
Step-by-step explanation:
Image result for True or False? If one of the angles of a triangle is obtuse, then the triangle is obtuse.
In an obtuse triangle, if one angle measures more than 90°, then the sum of the remaining two angles is less than 90°. Here, the triangle ABC is an obtuse triangle, as ∠A measures more than 90 degrees. ... Hence, if one angle of the triangle is obtuse, then the other two angles with always be acute.
