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3 years ago

Write the standard equation of a circle with center (-4, 4) and radius 1.5

Mathematics

1 answer:

Anna71 [15]3 years ago

4 0

Jfixijdjdkbfjdlnxbfjkenbxjkeb

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The problem is in the picture :)

frozen [14]

Answer:

Last option: 4

Step-by-step explanation:

The quadratic equation simplified: $x^2-4x=-\frac{7}{2}$ has the form:

$ax^2+bx=c$

In this case, you can identify that "a", "b" and "c" are:

$a=1\\b=-4\\c=-\frac{7}{2}$

To solve this quadratic equation by completing the square, Carlos should add $(\frac{b}{2})^2$ to both sides of the equation. This is:

$(\frac{-4}{2})^2=(-2)^2=4$

Then:

$x^2-4x+4=-\frac{7}{2}+4$

Therefore you can observe that the number he should add to both sides of the equation is: 4

8 0

3 years ago

Plsssssssss Help!!!!!!

Paraphin [41]

Answer:

Step-by-step explanation:

If shes out for 3 hours, times 2 by 3, 2x3 then they have a fee of 10 at the beginning then you add that on to 6 then you get 16. ( Sorry if I'

m wrong this is just how I was taught )

7 0

2 years ago

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Solve for x in terms of a,b, and c: ax - 3b =c. . 1: a(c+3b). 2: a (c-3b). 3: c-3/a. 4: c+3b/a

trapecia [35]

We have to solve x in terms of a, b and c:
a x - 3 b = c
a x - 3 b + 3 b = c + 3 b
a x = c + 3 b
x = ( c + 3 b ) : a
Answer:
4 ) ( c + 3 b ) / a

8 0

3 years ago

Solve for b help please

aliya0001 [1]

Answer:

-12=-4/3b

-12x4/3=b

-16=b

b=-16

4 0

3 years ago

Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

7 0

2 years ago

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