In this diagram, BAC~EDF. If the area of BAC=15 in2, what is the area of EDF? 4, 5.
1 answer:
Answer:
9.6 square inches.
Step-by-step explanation:
We are given that ΔBAC is similar to ΔEDF, and that the area of ΔBAC is 15 inches. And we want to determine the area of ΔDEF.
First, find the scale factor <em>k</em> from ΔBAC to ΔDEF:
Solve for the scale factor <em>k: </em>
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Recall that to scale areas, we square the scale factor.
In other words, since the scale factor for sides from ΔBAC to ΔDEF is 4/5, the scale factor for its area will be (4/5)² or 16/25.
Hence, the area of ΔEDF is:
In conclusion, the area of ΔEDF is 9.6 square inches.
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Yes. It is equivalent to 3. Combine like terms
The smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
What is the intermediate value theorem?
Intermediate value theorem is theorem about all possible y-value in between two known y-value.
x-intercepts
-x^2 + x + 2 = 0
x^2 - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1, x = 2
y intercepts
f(0) = -x^2 + x + 2
f(0) = -0^2 + 0 + 2
f(0) = 2
(Graph attached)
From the graph we know the smallest positive integer value that the intermediate value theorem guarantees a zero exists between 0 and a is 3
For proof, the zero exists when x = 2 and f(3) = -4 < 0 and f(0) = 2 > 0.
<em>Your question is not complete, but most probably your full questions was</em>
<em>Given the polynomial f(x)=− x 2 +x+2 , what is the smallest positive integer a such that the Intermediate Value Theorem guarantees a zero exists between 0 and a ?</em>
Thus, the smallest positive integer that the intermediate value theorem guarantees a zero exists between 0 and a is 3.
Learn more about intermediate value theorem here:
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