If f(x) = x2 + 1 and g(x) = x – 4, which value is equivalent to (f circle g) (10)?

Mathematics

1 answer:

Lemur [1.5K]3 years ago

4 0

Answer:

its 37

Step-by-step explanation:

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The sum of all three is 6x^2y + 2xy^2

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Bradley and Kelly are out flying kites at a park one afternoon. A model of Bradley and Kelly's kites are shown below on the coor

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The correct answer is:

C. They are similar because the corresponding sides of kites KELY and BRAD all have the relationship 2:1.

Using the distance formula,

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

the lengths of the sides of BRAD are:

$\text{BR}=\sqrt{(7-6)^2+(3-4)^2}=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\\\\\text{RA}=\sqrt{(6-3)^2+(4-3)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{AD}=\sqrt{(3-6)^2+(3-2)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{DB}=\sqrt{(6-7)^2+(2-3)^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}$

The lengths of the sides of KELY are:

$\text{KE}=\sqrt{(2-0)^2+(11-9)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\\\text{EL}=\sqrt{(0-2)^2+(9-3)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}\\\\\text{LY}=\sqrt{(2-4)^2+(3-9)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{40}=2\sqrt{10}\\\\\text{YK}=\sqrt{(4-2)^2+(9-11)^2}=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}$

Each side of KELY is twice the length of the corresponding side on BRAD. This makes the ratio of the sides 2:1 and the figures are similar.

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3 years ago

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mart [117]

This question can be solved using the Herons equation where
A = SQRT [s*(s-a)*(s-b)*(s-c)]

A = area of the triangle
a, b and c are the sides of the triangle

s = (a+b+c)/2

Since we are given the area, we can express "s" as a function of the third side c. This can be substituted in the original equation so as to obtain an expression to solve for the third side c

s = (4+10+c)/2 = (14+c)/2

using the solver function of the calculator or MS Excel, the third side is 7.21 units

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