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3 years ago

10

Find the polynomial equation with the given roots (precalc)

1 answer:

Alenkasestr [34]3 years ago

4 0

Theee point to under stand to how long but tee lol and then you have a good luck on the game you can have it to you

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Write the scientific notation for 2,460,000

Diano4ka-milaya [45]

It would be equalto 2.4 * 10^6

4 0

3 years ago

For what value(s) of k will the relation not be a function

skelet666 [1.2K]

We are given two relations

(a)

Relation (R)

$R=[((k-8.3+2.4k),-5),(-\frac{3}{4}k,4)]$

We know that

any relation can not be function when their inputs are same

so, we can set both x-values equal

and then we can solve for k

$k-8.3+2.4k=-\frac{3}{4} k$

$3.4k-8.3=-\frac{3}{4}k$

$3.4k\cdot \:10-8.3\cdot \:10=-\frac{3}{4}k\cdot \:10$

$4k-83=-\frac{15}{2}k$

$34k=-\frac{15}{2}k+83$

$\frac{83}{2}k=83$

$83k=166$

$k=2$ ............Answer

(b)

S = {(2−|k+1| , 4), (−6, 7)}

We know that

any relation can not be function when their inputs are same

so, we can set both x-values equal

and then we can solve for k

$2-|k+1|=-6$

$2-\left|k+1\right|-2=-6-2$

$-\left|k+1\right|=-8$

$\left|k+1\right|=8$

Since, this is absolute function

so, we can break it into two parts

$|f\left(k\right)|=a\quad \Rightarrow \:f\left(k\right)=-a\quad \mathrm{or}\quad \:f\left(k\right)=a$

$k+1=-8\quad \quad \mathrm{or}\quad \:\quad \:k+1=8$

we get

$k+1=-8\quad$

$k=-9$

$k+1=8\quad$

$k=7$

so,

$k=-9\quad \mathrm{or}\quad \:k=7$ ...............Answer

5 0

3 years ago

8y + 6 - 15y + 5 - 3

podryga [215]

Step-by-step explanation: im not really sure what it is but if you are working from this probplem do pemdas-

perthises,expoment,mulicaion,divishoin,addishoin, subbtraion

hope this helped! sorry if i spelt these wrong....

8 0

3 years ago

Sold rose bouquets and chocolate hearts for Valentin

GenaCL600 [577]

Answer:

do u want them all answered

Step-by-step explanation:

3 0

3 years ago

Find the center and radius of the circle whose equation is x 2 + y 2 + 10x − 1

Margaret [11]

we conclude that the center of the circle is the point (-5, 0).

<h3>How to find the center of the circle equation?</h3>

The equation of a circle with a center (a, b) and a radius R is given by:

$(x - a)^2 + (y - b)^2 = R^2$

Here we are given the equation:

$x^2 + y^2 + 10x - 1 = 0$

Completing squares, we get:

$x^2 + 10x + y^2 = 1$

Now we can add and subtract 25 to get:

$x^2 + 10x + 25 - 25 + y^2 = 1\\\\(x + 5)^2 -25 + y^2 = 1\\\\(x + 5)^2 + (y - 0)^2 = 26$

Comparing that with the general circle equation, we conclude that the center of the circle is the point (-5, 0).

If you want to learn more about circles:

brainly.com/question/1559324

#SPJ1

3 0

2 years ago