All categories

3 years ago

Factor the equation.

Mathematics

1 answer:

Kryger [21]3 years ago

5 0

Hi!

The fully factored form of this equation is (3x+2)²

You might be interested in

What is 21/49 in simplist from

ad-work [718]

If you divide 21/49 by 7 you get 3/7

8 0

3 years ago

LK is tangent to circle J at point K. Circle J is shown. Line segment J K is a radius. Line segment K L is a tangent that inters

Evgen [1.6K]

Answer:

$(B)r=\dfrac{85}{12}$

Step-by-step explanation:

Given a circle centre J

Let the radius of the circle =r

LK is tangent to circle J at point K

From the diagram attached

LX=6
Radius, XJ=JK=r
LK=11

Theorem: The angle between a tangent and a radius is 90 degrees.

By the theorem above, Triangle JLK forms a right triangle with LJ as the hypotenuse.

Using Pythagoras Theorem:

$(6+r)^2=r^2+11^2\\(6+r)(6+r)=r^2+121\\36+6r+6r+r^2=r^2+121\\12r=121-36\\12r=85\\r=\dfrac{85}{12}$

The length of the radius, $r=\dfrac{85}{12}$

8 0

3 years ago

Read 2 more answers

Can someone help please?

ratelena [41]

Answer: It is TRUE. A point in a linear equation can be the solution to said equation.

4 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

2 years ago

bag a has $3$ white marbles and $4$ black marbles. bag b has $6$ yellow marbles and $4$ blue marbles. bag c has $2$ yellow marbl

r-ruslan [8.4K]

Answer:

BALSS

Step-by-step explanation:

GURL Y DO YOU HAVE SO MANY MARBELS ???

4 0

1 year ago