Answer:
f(x) = (1/2)(x-2)^2(x+1)(x+2)
Step-by-step explanation:
You can determine this by looking at the zeroes of the graph. For any zero that goes through the x-axis, the power of that zero is odd. For any zero that that "bounces" from the x-axis, the power of that zero is even.
Starting from left to right, we can see that the first zero, -2, goes through the x-axis. That means (x+2) is raised to an odd power. The second zero, -1, also goes through, so (x+1) is raised to an odd power. The last zero, 2, bounces off the x-axis, so (x-2) is raised to an even power. The only functions that satisfy this criteria are function 1 and 2.
However, we are not done yet. We need to figure out which multiplier value (1/2, 1/4) is correct. To do this, we plug in 0 for x, since we know that the y-intercept is 4. When we plug in 0, we see that f(0) = 4 for the first function. Therefore, the first function is the answer.
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Answer:
54.4311
Step-by-step explanation:
divide the mass value by 2.205 (unit of conversion) :)
98.4252<span> inches is the exact so about 98 inches</span>
You save 5 cents per ounce by using the 40 ounce jar, because if you do $5.20 divided by 40 it equals $0.13 but if you do $2.70 divided by 15 it equals $0.18 which is 5 more cents the the 40 ounce

- Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>
- To find - <u>Area </u><u>of </u><u>trapezium</u>
Refer the figure attached ~
In the given figure ,
AB = 25 cm
BC = AD = 15 cm
CD = 13 cm
<u>Construction</u><u> </u><u>-</u>

Now , we can clearly see that AECD is a parallelogram !
AE = CD = 13 cm
Now ,

Now , In ∆ BCE ,

Now , by Heron's formula

Also ,

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

hope helpful :D