The LCD for the fractions 1/3, 3/4, 5/32, and 8/9 is

|6t−7|−8 ≥ 3 Please help me quick

Phantasy [73]

Hope the answer helps you!!!
:)

5 0

3 years ago

Help me on this one guys. I’m so bad at math as well as some other subjects

Svetradugi [14.3K]

Hey smart cookie ;)

So with this all we need to do is factor the expression.
Our answer is going to be 2x(16x^4+15)

Let me know if you have other questions!
Have a terrific day.
~Brooke❤️

7 0

4 years ago

The sum of two consecutive odd integers is at most 166. What are the greatest consecutive odd integers?

bagirrra123 [75]

. x + (x + 2)≤166
. 2x+2≤166
. x≤83 - 1
. x≤82
. x=81
. 81+(81+2)=164; 164<166
answer: 81 and 83

4 0

4 years ago

1.What are the zeros of the polynomial function?

Lorico [155]

Let's to the first example:

f(x) = x^2 + 9x + 20

Ussing the formula of basckara

a = 1
b = 9
c = 20

Delta = b^2 - 4ac

Delta = 9^2 - 4.(1).(20)

Delta = 81 - 80

Delta = 1

x = [ -b +/- √(Delta) ]/2a

Replacing the data:

x = [ -9 +/- √1 ]/2

x' = (-9 -1)/2 <=> - 5

Or

x" = (-9+1)/2 <=> - 4
_______________

Already the second example:

f(x) = x^2 -4x -60

Ussing the formula of basckara again

a = 1
b = -4
c = -60

Delta = b^2 -4ac

Delta = (-4)^2 -4.(1).(-60)

Delta = 16 + 240

Delta = 256

Then, following:

x = [ -b +/- √(Delta)]/2a

Replacing the information

x = [ -(-4) +/- √256 ]/2

x = [ 4 +/- 16]/2

x' = (4-16)/2 <=> -6

Or

x" = (4+16)/2 <=> 10
______________

Now we are going to the 3 example

x^2 + 24 = 14x

Isolating 14x , but changing the sinal positive to negative

x^2 - 14x + 24 = 0

Now we can to apply the formula of basckara

a = 1
b = -14
c = 24

Delta = b^2 -4ac

Delta = (-14)^2 -4.(1).(24)

Delta = 196 - 96

Delta = 100

Then we stayed with:

x = [ -b +/- √Delta ]/2a

x = [ -(-14) +/- √100 ]/2

We wiil have two possibilities

x' = ( 14 -10)/2 <=> 2

Or

x" = (14 +10)/2 <=> 12
________________

To the last example will be the same thing.

f(x) = x^2 - x -72

a = 1
b = -1
c = -72

Delta = b^2 -4ac

Delta = (-1)^2 -4(1).(-72)

Delta = 1 + 288

Delta = 289

Then we are going to stay:

x = [ -b +/- √Delta]/2a

x = [ -(-1) +/- √289]/2

x = ( 1 +/- 17)/2

We will have two roots

That's :

x = (1 - 17)/2 <=> -8

Or

x = (1+17)/2 <=> 9

Well, this would be your answers.

7 0

4 years ago

Three consecutive odd integers have a sum of 27. Find the integers.

jek_recluse [69]

$\sf \bf {\boxed {\mathbb {GIVEN:}}}$

Sum of three consecutive odd integers = $27$

$\sf \bf {\boxed {\mathbb {TO\:FIND:}}}$

The values of the three integers.

$\sf \bf {\boxed {\mathbb {SOLUTION:}}}$

$\sf\purple{The\:three\:consecutive \:odd\:integers\:are\:7,\:9\:and\:11.}$

$\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:\:EXPLANATION:}}}$

Let us assume the three consecutive odd integers to be $x$ , $(x+2)$ and $(x+4)$ .

As per the condition, we have

$Sum \: \: of \: \: the \: \: three \: \: consecutive \: \: odd \: \: integers = 27$

$➺ \: x + (x + 2) + (x + 4) = 27$

$➺ \: x + x + 2 + x + 4 = 27$

Now, collect the like terms.

$➺ \: (x + x + x) + (2 + 4) = 27$

$➺ \: 3x + 6 = 27$

$➺ \: 3x = 27 - 6$

$➺ \: 3x = 21$

$➺ \: x = \frac{21}{3} \\$

$➺ \: x = 7$

Therefore, the three consecutive odd integers whose sum is $27$ are $\boxed{ 7 }$ , $\boxed{ 9 }$ and $\boxed{ 11 }$ respectively.