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4 years ago

12

F(x)

1 answer:

sergey [27]4 years ago

8 0

Answer:

been to see if we have the

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Identify the vertex of y = x2 + 4x + 5.

Xelga [282]

ANSWER

The vertex is (-2,1)

EXPLANATION

We want to find the vertex of

$y = {x}^{2} + 4x + 5$

We complete the square to obtain,

$y = {x}^{2} + 4x + {(2})^{2} - {(2})^{2} + 5$

The first three terms forms a perfect square trinomial.

$y = {(x + 2})^{2} - 4 + 5$

The vertex form is

$y = {(x + 2})^{2} + 1$

This equation is in the form;

$y = a{(x -h})^{2} + k$

where (h,k)=(-2,1) is the vertex.

4 0

3 years ago

Read 2 more answers

if (2x minus 1) is exactly divisible by 6 X square + ax - 4 and b x square - 11x + 3, then the value of b square minus a square

alexdok [17]

Step-by-step explanation:

Given that:-

$(2x - 1)$

is exactly divisible by

$6 {x}^{2} + ax - 4$

and

$b {x}^{2} - 11x + 3$

We need to simply, place the value of x.

$2x - 1 = 0$

$2x = 1$

$x = \frac{1}{2}$

Now,

$6 {( \frac{1}{2} )}^{2} + \frac{a}{2} - 4 = 0$

$\frac{3 + a}{2} = 4$

$3 + a = 4 \times 2 = 8$

$a = 8 - 3$

$a = 5$

Now, to find b.

$b {( \frac{1}{2}) }^{2} - \frac{11}{2} + 3 = 0$

$\frac{b}{4} - \frac{11}{2} = - 3$

$\frac{b - 22}{4} = - 3$

$b - 22 = - 12$

$b = 10$

is the answer.

Now, we need

${b}^{2} - {a}^{2} + a - b$

$100 - 25 + 100 + 5$

$= 205 - 25$

$= 180$

is the answer.

Hope it helps :D

3 0

3 years ago

Help me please, i will brainliested ;)

Akimi4 [234]

The equation is y=-1/3-3

3 0

3 years ago

Barb is making a bead necklace. She strings 1 white bead, then 3 blue beads, then 1 white bead and so on. Write the numbers for

allsm [11]

You do it like this.

The mice are white beads and the gems are blue. It is 3:1 if you need the ratio.

5 0

4 years ago

Can someone help with 16 and 18

Effectus [21]

9514 1404 393

Answer:

16. k^5/j^3

18. 81y^20

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c) . . . . where b may be 0

(a^b)^c = a^(bc)

(ab)^c = (a^c)(b^c)

___

We can simplify the given expressions as follows.

<h3>16.</h3>

$\dfrac{k^6j^2}{kj^5}=k^{6-1}j^{2-5}=k^5j^{-3}=\boxed{\dfrac{k^5}{j^3}}$

__

<h3>18.</h3>

$(x^{-2})^2(3xy^5)^4=(x^{-2\cdot2})(3^4x^4y^{5\cdot4})=81x^{-4+4}y^{20}=\boxed{81y^{20}}$

6 0

3 years ago