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Arlecino [84]
4 years ago
7

Choose the constant term that completes the perfect square trinomial. y2 + y

Mathematics
2 answers:
Leviafan [203]4 years ago
7 0

In this question we will be using completing the square method.

We are given this expression:

y^{2} +y

Step 1:

Here we take half of coefficient of y , square it and then add and subtract it to the given expression :

y^{2} +y+(1/2)^{2} -(1/2)^{2}

y^{2}+y-1/4+1/4

So as we can see the constant term here is 1/4

horrorfan [7]4 years ago
5 0

The constant term is found by taking the coefficient of the singular variable (in this case, y), dividing it by 2, and squaring the result.

For this problem, the coefficient of y is 1. 1 divided by 2 is 1/2. (1/2)^2 = 1/4.

So, the constant term is 1/4.

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Need help. Please and thank u
Marina CMI [18]

Answer:

1)

(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12

Instead of multiplying 3 and 4 the student added them which gave them the answer of 7 when the correct answer is 12.

2)

a) x^2 + x - 2

b) x^2 - 7x + 10

c) 4x^2 - 1

d) x^2 + 10x + 25

Step-by-step explanation:

a. (x - 1)(x + 2)

x(x) - 1(x) + 2(x) - 1(2)

x^2 - 1x + 2x - 2

x^2 + x - 2

------------------------------------------

b. (x - 5)(x - 2)

x(x) - 5(x) - 2(x) - 5(-2)

x^2 - 5x - 2x + 10

x^2 - 7x + 10

------------------------------------------

c. (2x + 1)(2x - 1)

2x(2x) + 1(2x) - 1(2x) + 1(-1)

4x^2 + 2x - 2x - 1

4x^2 - 1

------------------------------------------

d. (x + 5)^2

(x + 5)(x + 5)

x(x) + 5(x) + 5(x) + 5(5)

x^2 + 5x + 5x + 25

x^2 + 10x + 25

7 0
3 years ago
Which system or equation is represented in the graph?
melamori03 [73]

Answer:

(A)\displaystyle  \begin{array}  {ccc}  \displaystyle y =  - 2 \\ x - 2y = 6\\  \end{array}

Step-by-step explanation:

<h3>Equation of the blue line:</h3>

since the line is horizontal and passes <u>-</u><u>2</u> the equation of the blue line should be

\displaystyle \large y =  - 2

<h3>Equation of the red line:</h3>

remember The form of equation of a line

\displaystyle \boxed{ \displaystyle  y = mx + b}

where m is slope of the line and b is the y-intercept we can clearly see that the red line crosses y-axis at (0,-3) therefore b=-3

to figure out m we can consider the following formula:

\displaystyle m =  \frac{ \Delta y}{ \Delta x}

from the graph we acquire ∆y=1 and ∆x=2

thus substitute:

\displaystyle m =  \frac{ 1}{ 2}

so we have figured out m and b

therefore our equation of blue line is

\displaystyle y =  \frac{1}{2} x - 3

our given options are in standard form so

move -3 to left hand side and change its sign:

\displaystyle y  + 3=  \frac{1}{2} x

cross multiplication:

\displaystyle 2y  + 6=  x

move 6 to right hand side and x to left hand side and change its sign

\displaystyle - x +  2y   =  -  6

multiply both sides by -1:

\displaystyle x  -   2y   =   6

hence, our system of linear equation is

\displaystyle \begin{cases}   \displaystyle y =  - 2 \\ x - 2y = 6\\ \end{cases}

3 0
3 years ago
100/225 square feet of your bakery will be used as your storefront, where customers come to visit. the rest will be for storage
Angelina_Jolie [31]
Okay, so the answer is 5/9. “Someone” was correct. The subtracting of 100 from 225 came from the 100 already being used out of that 225. If you subtract that, you’re left with how much space left. Place those back into the fraction and simplify and you get our answer.
5 0
4 years ago
Read 2 more answers
Khan Academy Volume of Sphere Answers!!!
agasfer [191]

Explanation:

The N stands for Pi so when you type in the answers to Khan Academy use the Pi symbol when theres an N! These are the common questions and common radius Khan Academy gives when they ask to find the radius of the circle.

Answers:

For 4 - 267.95  

For 1/4 - 1/48n  

For 1 - 4/3n    

For 9 - 972n

For 7 - 1372/3n

For 6 - 288n

For 3 - 36n

For 10 - 4000/3n

For 8 - 2048/3n

For 1/2 - 1/6n

For 5 - 500/3n

For 2 - 33.49

Step-by-step explanation:

<u><em>I did the go math so all of these rights and the majority of the answers are the questions Khan Academy will ask you to find the radius of the circle.</em></u>

5 0
2 years ago
Pre-Calculus - Systems of Equations with 3 Variables please show work/steps
Tatiana [17]

Answer:

x = 10 , y = -7 , z = 1

Step-by-step explanation:

Solve the following system:

{x - 3 z = 7 | (equation 1)

2 x + y - 2 z = 11 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y - 2 z = 11 | (equation 1)

x + 0 y - 3 z = 7 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y/2 - 2 z = 3/2 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x - y - 4 z = 3 | (equation 3)

Subtract 1/3 × (equation 2) from equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y - (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 16 × (equation 3) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y+0 z = 21 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 2 by -3:

{2 x + y - 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{2 x + 0 y - 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by 2:

{x+0 y+0 z = 10 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = 10 , y = -7 , z = 1

3 0
3 years ago
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