Question 75994
Problem #8
Solve:
radical (y+16) = 1
{{{sqrt(y+16)=1}}}
{{{(sqrt(y+16))^2=(1)^2}}}
{{{y+16=1}}}
{{{y+16-16=1-16}}}
{{{highlight(y=-15)}}}
Check to make sure it's not extraneaous:
{{{sqrt(-15+16)=1}}}
{{{sqrt(1)=1}}}
{{{1=1}}} The solution is valid.
:
Problem #9
Find the length x in the trianlge. Express your answer in simplified radical form.
The pythagorean theorem is {{{highlight(c^2=a^2+b^2)}}}, where c=hypotenuse and a and b are the legs of a right triangle
Hight 20cm
Legnth 15cm
Hypotenuse x
{{{x^2=(20)^2+(15)^2}}}
{{{x^2=400+225}}}
{{{x^2=625}}}
{{{sqrt(x^2)=sqrt(625)}}}
{{{highlight(x=25)}}}
The hypotenuse is 25 cm.
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Problem #10

Find the distance between (-7,7) and (2,-5)
The distance formula is based on the pythagorean theorem: {{{highlight(d=sqrt((x[2]-x[1])^2+(y[2]-y[1])^2))}}}
(x1,y1)=(-7,7) and (x2,y2)=(2,-5)
{{{d=sqrt((2-(-7))^2+(-5-7)^2)}}}
{{{d=sqrt((2+7)^2+(-5-7)^2)}}}
{{{d=sqrt((9)^2+(-12)^2)}}}
{{{d=sqrt(81+144)}}}
{{{d=sqrt(225)}}}
{{{highlight(d=15)}}}
:
Happy Calculating!!!!