An acute triangle has side lengths 21 cm, x cm, and 2x cm. if 21 is one of the shorter sides of the triangle, what is the greatest possible length of the longest side, rounded to the nearest tenth? 18.8 cm 24.2 cm 42.0 cm 72.7 cm

B. 24.2 cm

Step-by-step explanation:

We are given that,

A triangle has sides of measure x, 2x and 21 cm.

Using the Triangle Inequality Theorem, which states that, 'The sum of any two sides of the triangle is greater than the measure of the third side'.

As, 21 cm is the shorter side, we have the possible inequalities,

a). i.e. x < 21

b). i.e. i.e x > 7.

So, we have that, the range of the side of length x is

This gives us that,

The maximum value of the side of length x is less than 21 i.e. x < 21.

So, the maximum value of the side of length 2x is, 2x < 2×21 i.e. 2x < 42.

That is, the value of longest side is less than 42 (Since, 2x < 42).

Hence, the greatest possible length is 24.2 cm according to the options.

the answer would be d.

step-by-step explanation:

move all variables to the left side and all constants to the right side

(x−3)^2+(y−1)^2=16

this is the form of a circle. use this form to determine the center and radius of the circle.

( x − h ) 2 + ( y − k ) 2 = r 2

match the values in this circle to those of the standard form. the variable   r   represents the radius of the circle,   h   represents the x-offset from the origin, and   k   represents the y-offset from origin.

r = 4

h = 3

k = 1

the center of the circle is found at   ( h , k ) .

center:   ( 3 , 1 )

these values represent the important values for graphing and analyzing a circle.

center:   ( 3 , 1 ).

radius:   4

find the vertical and horizontal asymptotes of the function f x = 5 x − 1 . to find the vertical asymptote, equate the denominator to zero and solve for . since the degree of the polynomial in the numerator is less than that of the denominator, the horizontal asymptote is .

step-by-step explanation: