What are the perfect square trinomials?

Answers

Answer 1

Answer: A perfect square trinomial is

(square of one term) + (square of another term) + (double the product of the two terms)

Answer 2

Answer:

Let's look at the following trinomial.

X² - 14x + 49

(x - 7) (x - 7)

(x - 7)²

This trinomial would be classified as a perfect square trinomial because it factors as two identical binomials which is (x - 7)². This means that any trinomial that factors as two identical binomials is called a perfect square trinomial.

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Find the product and write it in lowest terms. 14/51x3/4

Answers

the answer to your question is : 14/51 * 3/4 = 7/34

14*3 = 42

------------------- = 42/207 (Then you reduce 42/204) and you get 42/204=7/34

51*4 = 204 (Hope this all makes sense :) )

The width,w, of a rectangular garden is X -2 the area of the garden is X^3-2X-4 what is an expression for the length of the garden? A. X^2-2x-2
B. X^2+2x-2
C. X^2-2x+2
D. X^2+2x+2

Answers

Answer:

D. $x^2+2x+2$

Step-by-step explanation:

We know that,

The area of a rectangle is,

A = l × b,

Where, l is the length of the rectangle,

w is the width of the rectangle,

Given,

$A = x^3-2x-4$

$w=(x-2)$

By substituting values,

$x^3-2x-4=(x-2)l$

$\implies l = (x^3-2x-4)/(x-2)=x^2+2x+2$ ( By long division shown below )

Hence, the length of the rectangular garden is $x^2+2x+2$

Option D is correct.

$A=lw \Rightarrow l=(A)/(w)$
A - area, l - length, w - width

$w=x-2 \nA=x^3-2x-4 \n \nl=(x^3-2x-4)/(x-2)=(x^3-2x^2+2x^2-4x+2x-4)/(x-2)=(x^2(x-2)+2x(x-2)+2(x-2))/(x-2)= \n=((x^2+2x+2)(x-2))/(x-2)=x^2+2x+2$

The answer is D. x²+2x+2.

Which of the following shows the correct factorization of x3 - 5x2 - 14x?A. x(x + 7)(x + 2)

B. x(x - 7)(x - 2)

C. x(x + 7)(x - 2)

D. x(x - 7)(x + 2)

Answers

So, just a note of formatting, to indicate something is an exponent of something else, use the carat (^) sign
for instance
x^3 means x cubed. x3 means 3x.
Anyways, though, let's factor.
You can factor out an x from each variable.
x(x^2-5x-14)
Now, to factor this, think of what will add to -5, and multiply to -14.
-7 and 2 work.
So, this factors out to
x(x-7)(x+2)
Choice D is the correct factorization.

Solve for x and y in the following set of equation 6x+5xy-5y=8
X+y=3

Please help this is important

Answers

STEP 1:

$6x+5xy-5y=8\n \n 6x+5y\left( x-1 \right) =8\n \n 5y\left( x-1 \right) =8-6x\n \n 5y\left( x-1 \right) =2\left( 4-3x \right) \n \n \frac { 1 }{ 5\left( x-1 \right) } \cdot 5y\left( x-1 \right) =2\left( 4-3x \right) \cdot \frac { 1 }{ 5\left( x-1 \right) } \n \n y=\frac { 2\left( 4-3x \right) }{ 5\left( x-1 \right) }$

STEP 2:

$x+y=3\n \n y=3-x$

STEP 3:

This means that...

$3-x=\frac { 2\left( 4-3x \right) }{ 5\left( x-1 \right) } \n \n \left\{ 5\left( x-1 \right) \right\} \left( 3-x \right) =2\left( 4-3x \right) \n \n \left( 5x-5 \right) \left( 3-x \right) =8-6x\n \n 15x-5{ x }^( 2 )-15+5x=8-6x\n \n 20x-5{ x }^( 2 )-15=8-6x\n \n 5{ x }^( 2 )+8-6x+15-20x=0\n \n 5{ x }^( 2 )-26x+23=0$

STEP 4:

Which means that...

$\n \frac { 1 }{ 5 } \cdot \left( 5{ x }^( 2 )-26x+23 \right) =0\cdot \frac { 1 }{ 5 } \n \n { x }^( 2 )-\frac { 26 }{ 5 } x+\frac { 23 }{ 5 } =0\n \n { x }^( 2 )-\frac { 26 }{ 5 } x=-\frac { 23 }{ 5 }$

Which means that...

$\ \n { \left( x-\frac { 13 }{ 5 } \right) }^( 2 )-{ \left( \frac { 13 }{ 5 } \right) }^( 2 )=-\frac { 23 }{ 5 } \n \n { \left( x-\frac { 13 }{ 5 } \right) }^( 2 )-\frac { 169 }{ 25 } =-\frac { 23 }{ 5 } \n \n { \left( x-\frac { 13 }{ 5 } \right) }^( 2 )=-\frac { 115 }{ 25 } +\frac { 169 }{ 25 } \n \n { \left( x-\frac { 13 }{ 5 } \right) }^( 2 )=\frac { 54 }{ 25 }$

Which means that...

$\n \n x-\frac { 13 }{ 5 } =\pm \frac { \sqrt { 54 } }{ 5 } =\pm \frac { 3\sqrt { 6 } }{ 5 } \n \n \therefore \quad x=\frac { 13 }{ 5 } \pm \frac { 3\sqrt { 6 } }{ 5 }$

STEP 5:

When:

$x=\frac { 13 }{ 5 } +\frac { 3\sqrt { 6 } }{ 5 }$

.........

$y=3-\left( \frac { 13 }{ 5 } +\frac { 3\sqrt { 6 } }{ 5 } \right) \n \n y=\frac { 15 }{ 5 } -\frac { 13 }{ 5 } -\frac { 3\sqrt { 6 } }{ 5 } \n \n y=\frac { 15-13-3\sqrt { 6 } }{ 5 } \n \n y=\frac { 2-3\sqrt { 6 } }{ 5 }$

STEP 6:

When :

$x=\frac { 13 }{ 5 } -\frac { 3\sqrt { 6 } }{ 5 }$

...........

$y=3-\left( \frac { 13 }{ 5 } -\frac { 3\sqrt { 6 } }{ 5 } \right) \n \n y=\frac { 15 }{ 5 } -\frac { 13 }{ 5 } +\frac { 3\sqrt { 6 } }{ 5 } \n \n y=\frac { 15-13+3\sqrt { 6 } }{ 5 } \n \n y=\frac { 2+3\sqrt { 6 } }{ 5 }$

Given h(x)=√5-x, the range of h, in inequality notation, is

Answers

assuming that the 5-x is under te sqrt sign

you cannot have sqrt of a negative
find where it equals zero
5-x=0
5=x

the limit is 5=x

x has to be greater than x
to infinity
D={x|x>5}

Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is . For one performance, advance tickets and same-day tickets were sold. The total amount paid for the tickets was . What was the price of each kind of ticket

Answers

Answer:

Advance tickets=$25

Same-day tickets=$15

Step-by-step explanation:

Complete question below:

Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is $ 40. For one performance, 25 advance tickets and 30 same-day tickets were sold. The total amount paid for the tickets was $1075 . What was the price of each kind of ticket?

Let

advance tickets=x

Same-day tickets=y

Combined cost of advance and same-day tickets=$40

It means,

x+y=40 Equ (1)

25 advance tickets and 30 same-day tickets=$1075

It means,

25x+30y=1075 Equ(2)

From (1)

x+y=40

x=40-y

Substitute x=40-y into (2)

25x+30y=1075

25(40-y)+30y=1075

1000-25y+30y=1075

5y=1075-1000

5y=75

Divide both sides by 5

5y/5=75/5

y=15

Recall,

x+y=40

x+15=40

x=40-15

=25

x=25

Advance tickets=$25

Same-day tickets=$15

Check

25x+30y=1075

25(25)+30(15)=1075

625+450=1075

1075=1075

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