<span><span>1. m2</span><span>(36<span>m4</span>+12<span>m2</span>+1)</span></span>2. Rewrite <span><span>m4</span><span>m4</span></span> as <span><span><span>(<span>m2</span>)</span>2</span><span><span>(<span>m2</span>)</span>2</span></span>.<span><span>3. m2</span><span>(36<span><span>(<span>m2</span>)</span>2</span>+12<span>m2</span>+1)</span></span>4. Let <span><span>u=<span>m2</span></span><span>u=<span>m2</span></span></span>. Substitute <span>uu</span> for all occurrences of <span><span>m2</span><span>m2</span></span>. =m2(36u2+12u+1)
5. Rewrite <span><span>36<span>u2</span></span><span>36<span>u2</span></span></span> as <span><span><span>(6u)</span>2</span><span><span>(6u)</span>2</span></span>.<span><span>5. m2</span><span>(<span><span>(6u)</span>2</span>+12u+1)</span></span>6. Rewrite <span>11</span> as <span>12</span><span><span>7. m2</span><span>(<span><span>(6u)</span>2</span>+12u+<span>12</span>)</span></span>8. Check the middle term by multiplying <span>2ab</span> and compare this result with the middle term in the original expression.<span><span>9. 2ab=2⋅<span>(6u)</span>⋅1</span><span>2ab=2⋅<span>(6u)</span>⋅1</span></span>10. Simplify. 2ab=12u
11. Factor using the perfect square trinomial rule <span><span><span>a2</span>+2ab+<span>b2</span>=<span><span>(a+b)</span>2</span></span><span><span>a2</span>+2ab+<span>b2</span>=<span><span>(a+b)</span>2</span></span></span>, where <span><span>a=6u</span><span>a=6u</span></span> and <span><span>b=1</span><span>b=1</span></span>.<span><span><span>12. m2</span><span><span>(6u+1)</span>2</span></span><span><span>m2</span><span><span>(6u+1)</span>2</span></span></span>13. Replace all occurrences of <span>uu</span> with <span><span>m2</span><span>m2</span></span>. m^2(6m^2+1)^2
Answer:
Range, Maximum, and y-intercept
Step-by-step explanation:
Range is the y-values of the function. Each of these functions has a range of (-inf, 10].
The maximum is the largest y-value that is part of the range, which is 10 for both functions.
y-intercept is where the function crosses the y-axis which is 10 for both functions.
Answer:
Kindly check explanation
Step-by-step explanation:
Suppose that the height s of a ball (in feet) at time t (in seconds) is given by the formula
s(t) = 64 - 16(t - 1)^2
t interval = 0 ≤ t ≤ 3
1) point A (from the graph)
11) Height of ball when it was released
Ball was released at t = 0
s(0) = 64 - 16(0 - 1)^2
= 64 - 16(-1)^2
= 64 - 16(1)
= 64 - 16
= 48 feets
111) point C ( from the graph)
IV) highest point of the ball is 64
Hence,
s(t) = 64 - 16(t - 1)^2
64= 64 - 16(t - 1)^2
16(t - 1)^2 = 64 - 64
16(t - 1)^2 = 0
16t^2 - 32t + 16 = 0
t^2 - 2t + 1 = 0
(t-1) = 0 (t-1) = 0
t = 1
V) Point G (from graph)
V1)
height = 0
s(t) = 64 - 16(t - 1)^2
0 = 64 - 16(t - 1)^2
16(t - 1)^2 = 64
(t - 1)^2 = 64/16
(t - 1)^2 = 4
(t - 1)² = 2²
t - 1 = 2
t = 2 + 1
t = 3
volume of paper bag= 16×5×20=1600cmcube
volume of box= 8×3×10= 240
space left = 1600-240= 1360cmcube
hope it helps