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2 years ago

How do you solve 4b^2+8b+-12=0. Also find the parabole.?

1 answer:

4 0

First you divide everything by 4 which will become b^2+2b-3=0
Then you can factor this which is
(b-1)(b+3)=0
B=1
B=-3 ... I think that's how you do it

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what is the following product of assume Y is greater than or equal to 0 square root y to the third Times Square Root why 2/3

sattari [20]

Step-by-step explanation:

$\sqrt{y3} \times \sqrt{y3}$

$\sqrt{y3 \times y3}$

$\sqrt{y3 + 3}$

$\sqrt{y6}$

4 0

2 years ago

Which statements are true? Check all that apply.

Dvinal [7]

Answer:

3is the answer because t

s apply

Step-by-step explanation:

3 0

1 year ago

Math algebra 2 show you’re work plz

LenKa [72]

9514 1404 393

Answer:

(t, u, w) = (1, -2, -2)

Step-by-step explanation:

A graphing calculator makes short work of this, giving the solution as ...

(t, u, w) = (1, -2, -2)

There are many ways to solve this "by hand." Here's one of them.

Add the first and third equations. Their sum is ...

-3t +4w = -11 . . . . . [eq4]

Add this to twice the second equation. That sum is ...

(-3t +4w) +2(-4t -2w) = (-11) +2(0)

-11t = -11

t = 1

Substituting this into the second equation gives ...

-4(1) -2w = 0

w +2 = 0 . . . . divide by -2

w = -2 . . . . add -2

Substituting for t in the third equation lets us find u.

2(1) -2u = 6

-1 +u = -3 . . . . . divide by -2

u = -2 . . . . add 1

The solution is (t, u, w) = (1, -2, -2).

5 0

2 years ago

Given the following definition, compute Q(5). Q(n) = 0 if n = 0 2 if n = 1 4 if n = 2 Q(n − 1) + Q(n − 2) + Q(n − 3) if n > 2

aalyn [17]

We have
$Q(0)=0\\
Q(1)=2\\
Q(2)=4$
and $Q(n)=Q(n-1)+Q(n-2)+Q(n-3)\mbox{ when } n\ \textgreater \ 2.$
So
$Q(3)=Q(2)+Q(1)+Q(0)=4+2+0=6\\
Q(4)=Q(3)+Q(2)+Q(1)=6+4+2=12\\
Q(5)=Q(4)+Q(3)+Q(2)=12+6+4=22.
$

3 0

2 years ago

2^5×5×11 and 2^3×5^2×7

IceJOKER [234]

Answer:

1. 1760

2. 1400

Step-by-step explanation:

2^5 = 32

32 × 5 = 160

160 × 11 = 1760

2^3 = 8

5^2 = 25

8 × 25 = 200

200 × 7 = 1400

4 0

2 years ago