Does this have infinite solutions or no solutions?

HELP! IM STUCK!

Readme [11.4K]

Answer:

$(x-1)^2+(y+4)^2=9$

Step-by-step explanation:

The equation of a circle is given by:

$(x-h)^2+(y-k)^2=r^2$

Where (h, k) is the center and r is the radius.

The center of the circle is (1, -4).

And the radius is 3.

Substitute:

$(x-(1))^2+(y-(-4))^2=(3)^2$

Simplify. Thus, our equation is:

$(x-1)^2+(y+4)^2=9$

3 0

3 years ago

An integer is chosen at random from 1 to 50 inclusive. Find the probability that the chosen integer is divisible by 3

kompoz [17]

If an integer is chosen between 1 and 50 inclusive, you have 50 numbers total to deal with, so 50 is in the denominator of our ratio (fraction). All the numbers divisible by 3 in that interval total 16 numbers. So the ratio would be 16/50 for a percentage of 32%

6 0

3 years ago

Read 2 more answers

Six identical squares are cut from the corners and edges of an 80 cm by 50 cm cardboard rectangle. the remaining piece is folded

Anna11 [10]

Check the picture.

let the length of a side of each of the squares removed be x.

The box formed will have dimensions: 80-2x, 50-2x, x(the height)

So the volume can be expressed as a function of x as follows:

f(x)=(80-2x)(50-2x)x= $[4000-160x-100x+4 x^{2} ]x=(4 x^{2}-260x+4000)x$

so $f(x)=4 x^{3}-260x^{2}+4000x$

the solutions of f'(x)=0 gives the inflection points, so the candidates for maxima points,

$f'(x)=12x^{2}-520 x +4000=0$

solving the quadratic equation, either by a calculator, graphing software, or by other algebraic methods as the discriminant formula, we find the solutions

x=10 and x=33.333

plug in f(x) these values to see which greater:

$f(10)=(80-20)(50-20)10=60*30*10=18000$ cm cubed

$f(33.333)=(80-66.666)(50-66.666)33.333=$ which is negative because (50-66.666)<0

Answer: 18000 cm cubed

5 0

3 years ago

Read 2 more answers

For each part, give a relation that satisfies the condition. a. Reflexive and symmetric but not transitive b. Reflexive and tran

Vesnalui [34]

Answer:

For the set X = {a, b, c}, the following three relations satisfy the required conditions in (a), (b) and (c) respectively.

(a) R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)} is reflexive and symmetric but not transitive .

(b) R = {(a, a), (b, b), (c, c), (a, b)} is reflexive and transitive but not symmetric .

(c) R = {(a,a), (a, b), (b, a)} is symmetric and transitive but not reflexive .

Step-by-step explanation:

Before, we go on to check these relations for the desired properties, let us define what it means for a relation to be reflexive, symmetric or transitive.

Given a relation R on a set X,

R is said to be reflexive if for every $a \in X, (a,a) \in R$ .