I: 12x-5y=0
II:(x+12)^2+(y-5)^2=169
with I:
12x=5y
x=(5/12)y
-> substitute x in II:
((5/12)y+12)^2+(y-5)^2=169
(25/144)y^2+10y+144+y^2-10y+25=169
(25/144)y^2+y^2+10y-10y+144+25=169
(25/144)y^2+y^2+144+25=169
(25/144)y^2+y^2+169=169
(25/144)y^2+y^2=0
y^2=0
y=0
insert into I:
12x=0
x=0
-> only intersection is at (0,0) = option B
Answer:
(-1/2, 27/2)
Step-by-step explanation:
Midpoint formula: (x^1 +x^2/ 2, y^1 +y^2/ 2)
(-4 +3 /2, 12+15/ 2) -> (-1/2, 27/2)
Answer:
<em>Well, Your answer will be is </em><em>D. 100 feet squared. </em><em>Because, </em>
<em>1. Multiply 10x6 and 10x4.
</em>
<em>
</em>
<em>2. Add 60 and 40, the results from the previous step.
</em>
<em>
</em>
<em>3. You get 100, or 100 feet squared. </em><em>Good Luck!</em>
<em>
</em>
<em />
Answer:
6 units is the answer im pretty sure
The value of f(a)=4-2a+6
, f(a+h) is 
, [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6
.
Given a function f(x)=4-2x+6.
We are told to find out the value of f(a), f(a+h) and [f(a+h)-f(a)]/hwhere h≠0.
Function is like a relationship between two or more variables expressed in equal to form.The value which we entered in the function is known as domain and the value which we get after entering the values is known as codomain or range.
f(a)=4-2a+6 (By just putting x=a).
f(a+h)==
=4-2a-2h+6(
)
=4-2a-2h+6
=
[f(a+h)-f(a)]/h=[-(4-2a+6 )]/h
=
=
=6h+12a-2.
Hence the value of function f(a)=4-2a+6, f(a+h) is , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6.
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