All categories

3 years ago

Use the “complete the square” method to rewrite the equation ax 2 + bx + c = 0 in the form ( x – p ) 2 = q .

2 answers:

5 0

X^2 + (b/a)x + c/a = 0
x^2 + (b/a)x + c/a = 0
x^2 + 2(b/a)x + c/a - b/a = 0
x^2 + 2(b/a)x = b/a - c/a
x^2 + 2(b/a)x + (b/a)^2 = b/a - c/a + (b/a)^2
(x + b/a)^2 = b/a - c/a + (b/a)^2
p = -b/a
q = b/a - c/a + (b/a)^2

mel-nik [20]3 years ago

3 0

Ax2 + bx + c = 0 = a(x+d)^2 + e = 0

You might be interested in

ONLY ANSWER IF YOU KNOW IT PLS ILL GIVE THE CORRECT BRAINLIEST!

Bas_tet [7]

Answer:

y=5

Step-by-step explanation

just sub in for x

Y=10-5

7 0

3 years ago

A pet store holds training classes for dogs. The store has two different training programs. The first program charges a $35 memb

GrogVix [38]

The programs cost the same for 7 classes.

Given, no. of classes is represented by c, no. of dogs is represented by d, no. of hours is represented by h, the no. of programs is represented by p and the total cost is represented by t.

The first program charges $35 membership fee and $5 for each class.

The second program charges only $10 for each class.

Now the total cost say $t_{1}$ , for first program can be written in equation form as,

$t_{1} =35+5c$ , here c is the no. of classes.

Similarly, the total cost say $t_{2}$ , for second program can be written in equation form as,

$t_{2} =10c$ .

We have to find out the no. of classes for which the two programs cost the same.

according to the question,

$t_{1} =t_{2}$

$35+5c=10c\\10c-5c=35\\5c=35\\c=7$

Hence for 7 of classes the programs cost the same.

For more details follow the link:

brainly.com/question/4042361

3 0

2 years ago

Read 2 more answers

The graph shows the relationship between the pounds of candy bought at a candy shop and the total cost, in dollars.

Vesna [10]

C I took the test and got it right

6 0

2 years ago

In a math class, 12 people reported their grades on the last test. Here are their responses:

Murrr4er [49]

Answer:

(a) 23, 35, 28, 33, 5, 12, 40, 25, 20, 18, 1, 16

1, 5, 12, 16, 18, 20 23, 25, 28, 33, 35, 40

n = 12 M = (20 + 23)/ 2 = 21.5

Q1 = (12 + 16) / 2 = 14 Q3 = (28 + 33) /2 = 30.5

(b) 22, 33, 25, 28, 5, 12, 35, 23, 20, 18, 1, 40, 16

1, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 33, 35, 40

n = 13 M = 22 Q1 = 14 Q3 = 30.5

1, 3, 5, 12, 16, 18, 20, 22, 23, 25, 28, 33, 35, 40

n = 14 M = (20 + 22)/2 = 21 Q1 = 12 Q3 = 28

(d) 20, 28, 23, 25, 3, 5, 30, 22, 18, 40, 16, 35, 1, 33, 12

1, 3, 5, 12, 16, 18, 20 | 22 | 23, 25, 28, 30, 33, 35, 40

n = 15 M = 22 Q1 = 12 Q3 = 30

hope this helps :)

5 0

3 years ago

A gumball machine has different flavors sour apple,grape,orange and cherry . There are six of each flavor. $.50 are put in the m

tamaranim1 [39]

Answer:

30/552

Step-by-step explanation:

In order to solve this problem you need to multiply the probability of getting grape for the first gumball with the probability of getting grape for the second gumball. Since there are 6 grape gumballs and a total of 24 gumballs (6*4). Then the probability of getting grape for the firs one is

$\frac{6}{24}$

Now there are only 5 grape gumballs available and one less in the total supply, therefore the probability of getting grape in the second try is

$\frac{5}{23}$

Finally we multiply them together to find the probability of getting two grapes in a row.

$\frac{6*5}{24*23}$ = $\frac{30}{552}$

4 0

3 years ago