SOMEONE PLEASE HELP ASAP!!! WILL GIVE BRAINLIEST!!!!!!!!!!!!

Mathematics

1 answer:

shtirl [24]3 years ago

5 0

There are actually infinite possibilities,

Here are a few, 9/20, 17/40, 19/40

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Find all the points, if any, where the graph of 12x-5y=0 intersects (x+12)^2+(y-5)^2=169.

WITCHER [35]

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

6 0

3 years ago

Read 2 more answers

What is 6-2p=p+9 ?Because I don't know it

nikklg [1K]

Hey there!

Firstly, you have to $\bold{simplify}$ each of the sides of your equation
$\bold{6-2p-p+9\downarrow}$
$\bold{6+(-2p)=p+9\downarrow}$
$\rightarrow\bold{-2p+6=p+9}$
Now that you have that portion out of the way, we could move on to our second step... which is to $\bold{subtract}$ by the value of $\bold{p}$ on each of your sides
$\bold{-2p+6-p-p+9-p\downarrow}$
$\rightarrow\bold{-3p+6-9}$ (you get this because we solve the particular equation we added above)
Thirdly, you have to $\bold{subtract}$ by the number $\bold{6}$ on each of your sides
$\bold{-3p+6-6=9-6\downarrow}$
$\bold{Cancel:6-6}$ because it equals to $\bold{0}$
$\bold{Keep:9-6}$ because it brings us closer and closer to the answer of $\bold{p}$
Fourthly, we have to $\bold{divide}$ by the number $\bold{-3}$ on each of your sides
$\bold{\frac{-3p}{-3}=\frac{3}{-3}}$
$\boxed{\boxed{\bold{Answer:p=-1}}}$ $\checkmark$