Question

Updated need help again,thanks.

Answer 1

The diameter? These are both equations for circles. Do you know how to find the radius from the equation?

Once you know the radius, just multiply it by 2 to get the diameter.

Answer 2

Standard form equation for a circle with center (h,k) and radius r is:

\(\large(x-h)^{2}+(y-k)^{2}=r^2\)

Answer 3

this is one equation not 2 and i don't know to solve nothing about this,and i need your help to tell me how to solve it i mean i really want to see it step by step

Answer 4

I think @niku2k should define the variable d, ¿what is d? (diameter, distance between centers, etc)

Answer 5

it is distance, btw i would really apreciate if somone can help me t this one also
the rproblem in image nnumber 17

Answer 6

The solution you propose in problem 17 is ok, do you know how to reach this solution?

Answer 7

In the first problem you propose, the centers of the circunferences are,
\[P_1=(3,-2), \ \ P_2=(-4,-3)\]
To find d you must find the distance between this points,
\[d=\sqrt{(-3-(-2))^2+(-4-3)^2}=\sqrt{50}=5\sqrt{2}\]

Answer 8

OK; sorry - you said "find d" and I assumed that, since these were circle equations, "d" meant "diameter". Had you said "find the distance between the centers of the circles", I would have answered differently, lol. :)

Answer 9

thank you john so much,and the other question i just know the rezultat but not how to solve

Answer 10

You can try extract common factor from the left side (2^x) and from the right side (3^x).

Answer 11

If your question on #17 is how to arrive at x=4 (in a rigorous way - I'm assuming you did some guessing and checking to get it here)... then:

\(\large 2^{x-1}-2^{x-3}=3^{x-2}-3^{x-3}\)

By rules of exponents:
\(\large 2^{x-1}(1-2^{-2})=3^{x-1}(3^{-1}-3^{-2})\)

This can be solved for x by rearranging, and putting both sides in the same base. :)

Answer 12

\(\large \dfrac{2^{x-1}}{3^{x-1}}=\dfrac{(3^{-1}-3^{-2})}{(1-2^{-2})}\)

See if you can get it from there... it's actually quite a slick little problem, I like it! :)

Answer 13

Although, I would choose a little different way,
\[2^x(2^{-1}-2^{-3})=3^x(3^{-2}-3^{-3})\]\[2^x(1/2-1/8)=3^x(1/9-1/27)\]
\[2^x\frac{3}{8}=3^x\frac{2}{27}\Rightarrow \frac{2^x}{3^x}=\frac{16}{81}\]
And then obtain x.

Answer 14

Yes, same idea, and I like pulling out the power x better, too....not sure why I went for x-1, for some reason my brain just went there first, lol. :) Both ways work though..... gotta love math! :)

I was having connection problems and didn't see your suggestion about 2^x and 3^x until I had posted my method, lol. :)

Answer 15

Thank you guys for helping me,i had q question before 2 hours ago but nobody helped me can i ask you one more time ,and i hope its not to much

here it is -x+3y+λ(x+2y+4)=0 the straight which passes through point (-1,1)is i hope it is understandable

Answer 16

Could you put the original question in your language (with an image)?

Answer 17

yes i guess hold on a minute

Answer 18

@DebbieG is answering so may be it's no needed to post it. ;)

Answer 19

No, I don't quite understand either... I *think* maybe we want λ such that:

-x+3y+λ(x+2y+4)=0

is a line passing through (-1,1)? I'm trying to see if that works out... lol

Answer 20

I realize after you were writint that may be solution should be in this way. Simpliy plug in the point they give you (x,y)=(-1,1) and obtain the value of the parameter,
\[1+3+\lambda(-1+2+4)=0\]
Could you finish it?

Answer 21

One you obtain the parameter, you should susbtitute in the original equation, and obtain the answer of the problem.

Answer 22

Yup, that does the trick. :)

Answer 23

okay guys thanks so much for helping me and i made a photo about the question,but i really need jsut a little bit mroe help by telling me how to solve some question in the images question are with red circle thanks so much

Answer 24

Only to complete,
\[-x+3y+(-4/5)(x+2y+4)=0\Rightarrow - 9 x + 7 y-16=0\]
For the other 2,
1) I think in the problem says "find the value a that makes both lines perpendicular". In order to be perpendicular it must be,
\[3a\cdot5-4\cdot6=0\]
and solve for a.

Answer 25

2) You need some trigonometric relations,
\[1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\]
So,
\[\sin\alpha=\frac{1}{\sqrt{1+\cot^2\alpha}}\]
Substitute the value they give you, and then you'll find the solution.

Answer 26

6) Find the center of the circles,
For the frist one (-2,2) and for the second (0,-3), completing squares. Then find the equation through these points,
\[y-y_0=\frac{y_1-y_0}{x_1-x_0}(x-x_0)\]
where,
\[(x_0,y_0)=(-2,2)\]\[(x_1,y_1)=(0,-3)\]