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3 years ago

X2 −2x + y2 − 6y = 26 what is the radius

1 answer:

4 0

$r > 0\\\\(x-a)^2+(y-b)^2=r^2\\\\x^2-2x+y^2-6y=26\\\\ (x-1)^2-1+(y-3)^2-9=26\\\\ (x-1)^2+(y-3)^2-10=26 \ |+10\\\\(x-1)^2+(y-3)^2=36\\\\ r^2=36 \ \Rightarrow \ r=6$

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!! Factor. 11x^2 + 35x + 6

fredd [130]

Hey there!!

Given equation :

11x² + 35x + 6

Now let's write 35x as 33x and 2x

Then the equation would become :

... 11x² + 33x + 2x + 6

... Now, let's take the common terms 11x² + 33x and 2x + 6

It would become :

... 11x ( x + 3 ) + 2 ( x + 3 )

... Now, we will write this as :

... ( 11x + 2 ) ( x + 3 )

Hence, this is as the answer...

Hope my answer helps!!

6 0

3 years ago

The intersection of two planes is a what

photoshop1234 [79]

The intersection of two planes is a line. :) Hope it works.

7 0

4 years ago

Flaky pastry can be made using flour and fat in the ratio 4 : 3. Jake makes some flaky pastry using 90 grams of fat. What weight

lubasha [3.4K]

Answer:

He use 120 grams of flour.

Step-by-step explanation:

Given:

Flaky pastry can be made using flour and fat in the ratio 4 : 3. Jake makes some flaky pastry using 90 grams of fat.

Now, to find the weight of flour he use in making flaky pastry.

Let the weight of flour be $x.$

Weight of fat = 90 grams.

The ratio of flour and fat used in making flaky pastry is 4 : 3.

As, 4 is equivalent to 3.

Thus, $x$ is equivalent to 90.

Now, to solve using cross multiplication method:

$\frac{4}{3} =\frac{x}{90} \\\\By\ cross\ multiplying\ we\ get:\\\\360=3x\\\\Dividing\ both\ sides\ by\ 3\ we\ get:\\\\120=x\\\\x=120\ grams.$

Therefore, he use 120 grams of flour.

4 0

3 years ago

Find the inverse laplace transform of: (2 s + 4) / (s - 3)^3

Serhud [2]

Answer:

$e^{3t}(2t+5t^{2})$

Step-by-step explanation:

$L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=$

Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3

Translation theorem: $L^{1} [F(s-a)=L^{-1}[F(s)|_{s \to s-a}\\ L^{1} [F(s-a)=e^{at} f(t)$

$L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=e^{3t} L^{-1}[\frac{2(s+3)+4}{s^{3}} ]$

Separate the fraction in a sum:

$e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])$