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3 years ago

Question: write your quadratic equation y=x^2-6x+7 in vertex form.

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

7 0

y=(x-3)^2-2

find the h value using h=-b/2a
plug in the value (3) into the equation to solve for k
use the values to write the equation in vertex form which is y=a(x-h)^2+k

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At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4. During the year, 10 juniors and twice

Lady bird [3.3K]

Answer: E. 120

The number of seniors were there in high school X at the beginning of the year = 120

Step-by-step explanation:

Given : At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4.

Let the number of juniors be 3x and the number of seniors be 4x.

Since , during the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school X.

i.e. Number of juniors at the end of the year = 3x-10

Number of seniors at the end of the yea = 4x-2(10)=4x-20

At the end of the year, the ratio of juniors to seniors was 4 to 5.

$\Rightarrow\dfrac{3x-10}{4x-20}=\dfrac{4}{5}$

$\Rightarrow5(3x-10)=4(4x-20)$

$\Rightarrow15x-50=16x-80$

$\Rightarrow16x-15x=80-50$

$\Rightarrow x=30$

The number of seniors were there in high school X at the beginning of the year = 4(30)=120

Hence, the correct answer is E. 120 .

6 0

3 years ago

Please help me i need help points and brainly

snow_lady [41]

The values go up by 2. So your answer would be 20.

4 0

3 years ago

Read 2 more answers

How to solve questions

zhuklara [117]

Too shaky, cant read

7 0

3 years ago

Match each expression with its equivalent expression.

Klio2033 [76]

Answer:

a -> 3

b -> 4

c -> 1

d -> 5

e -> 2

Step-by-step explanation:

We apply the properties to solve this question.

a. √4x^2y^4

$\sqrt{4x^2y^4} = \sqrt{4}\sqrt{x^2}\sqrt{y^4} = 2xy^2$

So a -> 3

b. √8x^2y

$\sqrt{8x^2y} = \sqrt{8}\sqrt{x^2}\sqrt{y} = \sqrt{4*2}x\sqrt{y} = \sqrt{4}\sqrt{2}x\sqrt{y} = 2x\sqrt{2}\sqrt{y} = 2x\sqrt{2y}$

So b -> 4

c. √4x^2y

$\sqrt{4x^2y} = \sqrt{4}\sqrt{x^2}\sqrt{y} = 2x\sqrt{y}$

So c -> 1

d. √16xy^2

$\sqrt{16xy^2} = \sqrt{16}\sqrt{x}\sqrt{y^2} = 4\sqrt{x}y = 4y\sqrt{x}$

So d -> 5

e. √8xy^2

$\sqrt{8xy^2} = \sqrt{8}\sqrt{x}\sqrt{y^2} = \sqrt{4*2}\sqrt{x}y = \sqrt{4}\sqrt{2}\sqrt{x}y = 2y\sqrt{2}\sqrt{x} = 2y\sqrt{2x}$

So e -> 2

5 0

3 years ago

Which is Not equal to thr absolute value of -5?

USPshnik [31]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers