<span>A)Both m(x) and p(x) cross the x-axis at 7.
B)Both m(x) and p(x) cross the y-axis at 7.
C)Both m(x) and p(x) have the same output value at x = 7.
D)Both m(x) and p(x) have a maximum or minimum value at x = 7.
m(x) = p(x) at x = 7
</span><span>
True statement about x = 7.
C)Both m(x) and p(x) have the same output value at x = 7. </span><span>
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1/3( 4 - 5x - 1)
= (1/3)(4 + - 5x + - 1)
= (1/3)( 4) + (1/3)(- 5x) + (1/3)( - 1)
= 4/3 + - 5x/3 + - 1/3
= - 5x/3 + 1 (Decimal: - 1.666667x + 1)
Hope that helps!!!!
Answer:
They have both root common .
Step-by-step explanation:
it filled up half the circle (up to the center point) - if we had a full circle. but a little bit is cut off (below AB).
what we see is that the shaded area is the sum of the area of the triangle AOB and 2 equally sized circle segment areas left and right of AOB.
since we are dealing with a half-circle, we have 180° in total. 120° are taken by AOB, so, that leaves us with 180-120 = 60° for both circle segments (so, one has an angle of 30°).
and 2×30° = 1×60°, so we can calculate the area of one 60° segment instead of two 30° segments.
AOB is an isoceles triangle (the legs are equally long, and therefore also the 2 side angles are equal).
the area of this triangle AOB is
1/2 × a × b × sin(C) = 1/2 × 3 × 3 × sin(120) =
= 3.897114317... m²
a circle segment area of 60° is 60/360 = 1/6 of the full circle area (as a full circle = 360°).
so, it's area is
pi×r² × 1/6 = pi×3²/6 = pi×3/2 = 4.71238898... m²
so, the total area of the shaded area is
3.897114317... m² + 4.71238898... m² =
= 8.609503297... m²
Answer:
B
Step-by-step explanation:
- 2 | 2 2 - 2 4
↓ - 4 4 - 4
--------------------------
2 - 2 2 0 ← remainder
Since the remainder = 0 then (x + 2) is a factor
quotient = 2x² - 2x + 2
