MATHEMATICS HIGH SCHOOL

A student is deriving the quadratic formula. Her first two steps are shown.Step 1: –c = ax2 + bx

Step 2: -c = a[x^2+b/ax]

Which best explains or justifies Step 2?

division property of equality

factoring the binomial

completing the square

subtraction property of equality

Answers

Answer 1

Answer:

The polynomial in its standard form is:

The student's first step is:

The student's second step is:

The student's second step is to factor the remaining binomial on the right side of the equation.

In this case, the student made a common factor of the letter a.

Answer:

The sentence that best explains or justifies Step 2 is:

factoring the binomial

Answer 2

Answer:

Answer:

factotoring

Step-by-step explanation:

Related Questions

Which ordered pair is a solution to the system of equations? Use the graph.x+2y=4 (line a) -3x+2y=-4 (line b) A. (2,-1) B. (-2,1) C. (2,1) D. (-2, -1)
3/8 divided by 1/2 =
Solve the following system of equations for x: 4x + y = 11 x + y = 2
What’s the perimeter and area
If the volume for 3-D polyhedra A is 200 cm3 and the volume for 3-D polyhedral B is 800 cm3, how many times bigger is the volume of pyramid B than pyramid A? A. 0.4% B. 4% C. 40% D. 400%

Solve the system of equations 3x+2y=14 2x-4y=4

Answers

3x + 2y = 14....multiply by 2
2x - 4y = 4
----------------
6x + 4y = 28 (result of multiplying by 2)
2x - 4y = 4
----------------add
8x = 32
x = 32/8
x = 4

2x - 4y = 4
2(4) - 4y = 4
8 - 4y = 4
-4y = 4 - 8
-4y = -4
y = -4/-4
y = 1

solution is (4,1)

Jorge baked cookies for his math class's end-of-year party. There are 28 people in Jorge's math class including Jorge and his teacher. Jorge baked enough cookies for everyone to get 3 cookies each. How many cookies did Jorge bake?

Answers

You have to multiply the number of people times the number of cookies they are each supposed to get

So, 28 x 3 = 84 cookies

For the function f(x) = -2(x + 3)^2 -1, identify the vertex, domain, and range. A.) The vertex is (3, -1), the domain is all real numbers, and the range is y ≥ -1. B.) The vertex is (3, -1), the domain is all real numbers, and the range is y ≤ -1. C.) The vertex is (-3, -1), the domain is all real numbers, and the range is y ≤ -1. D.) The vertex is (-3, -1), the domain is all real numbers, and the range is y ≥ -1

Answers

ok so domain is all allowed numbers look at deonomenator and don't allow any numbers that will make it zero none that means domain=all real numbers vertex in form y=a(x-h)^2+k (h,k)=vertex we have y=-2(x+3)^2-1 h=-3 k=-1 vertex=(-3,-1) range the vertex opens down so max is y=-1 y≤-1 domain is all real vertex is (-3,-1) y≤-1 C

Find the GCF of 44j5k4 and 121j2k6.Find the GCF of 44 and 121.

Find the GCF of j5 and j2.

Find the GCF of k4 and k6.
What is the GCF of 44j5k4 and 121j2k6?
4j3k2
4j2k4
11j3k2
11j2k4 (this is the correct one)

Answers

Answer:

D. $11j^2k^4$

Step-by-step explanation:

We are asked to find the GCF of $44j^5k^4\text{ and }121j^2k^6$ .

Since we know that GCF of two numbers is the greatest number that is a factor of both of them.

First of all we will GCF of 44 and 121.

Factors of 44 are: 1, 2, 4, 11, 22, 44.

Factors of 121 are: 1, 11, 11, 121.

We can see that greatest common factor of 44 and 121 is 11.

Now let us find GCF of $j^5\text{ and }j^2$ .

Factors of $j^5$ are: $j*j*j*j*j$

Factors of $j^2$ are: $j*j$

We can see that greatest common factor of $j^5\text{ and }j^2$ is $j*j=j^2$ .

Now let us find GCF of $k^4\text{ and }k^6$ .

Factors of $k^4$ are: $k*k*k*k$

Factors of $k^6$ are: $k*k*k*k*k*k$

We can see that greatest common factor of $k^4\text{ and }k^6$ is $k*k*k*k=k^4$ .

Upon combining our all GCFs we will get,

$11j^2k^4$

Therefore, GCF of $44j^5k^4\text{ and }121j^2k^6$ is $11j^2k^4$ and option D is the correct choice.

its 11j2k4 because well i just know trust me my dude

Which of the following points is a solution of y > |x| + 5?a. (0, 5)
b. (1, 7)
c. (7, 1)

Answers

You can do this one simply just by plugging in the values of x and y for each answer.
Note: (0,5) is just a way to right x=0 and y=5

Try answer A (0,5)
$y\ \textgreater \ |x|+5$
$5\ \textgreater \ |0|+5$
$5\ \textgreater \ 0+5$
$5\ \textgreater \ 5$
This is not true. 5 is not bigger than 5 so A is not the answer

Try answer B (1,7)
$y\ \textgreater \ |x|+5$
$7\ \textgreater \ |1|+5$
$7\ \textgreater \ 1+5$
$7\ \textgreater \ 6$
This is true so answer B works!

Might as well try answer C (7,1)
$y\ \textgreater \ |x|+5$
$1\ \textgreater \ |7|+5$
$1\ \textgreater \ 7+5$
$1\ \textgreater \ 12$
1 is obviously not greater than 12 so C does not work

The answer is B (1,7)

If A(1/3.5, 5) and B(10,39)..
Find the slope