Geometry math question no Guessing and Please show work thank you
2 answers:
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Answer:
4
Step-by-step explanation:
The general equation of a circle with center (a, b) and radius r is given by the equation;
The constant in the right hand side of the equation is simply the square of the radius;
We have been given the following equation;
(x-3)^2 + (y+1)^2 = 16
Comparing this with the general equation above;
Answer:
x = 12
Step-by-step explanation:
The midsegment VY is half the sum of the parallel bases , that is
VY = , then
3x + 18 = (x + 96) ← multiply both sides by 2 to clear the fraction
6x + 36 = x + 96 ( subtract x from both sides )
5x + 36 = 96 ( subtract 36 from both sides )
5x = 60 ( divide both sides by 5 )
x = 12
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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