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3 years ago

13

Question 2!! Someone help ASAP!! I need help!! I don’t Understand

1 answer:

Inessa05 [86]3 years ago

8 0

I don’t understand Exactly Either but you could ask the internet text someone in your family to

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How many 1/12pieces are in 3/3?

wel

Answer:

plz ion know what is the awnser to mine

Step-by-step explanation:

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2 years ago

How do you simplify each expression 3-5w+12+2w-3

Wittaler [7]

3-5w+12+2w-3
(3+12-3)=12
(-5w+2w)=-3w
-3w+12

5 0

2 years ago

Krutika and Gavin have played 41 tennis matches.

Neko [114]

Answer:

P(Kritika)= 27/41

P(Gavin)=1-(27/41)

= (41-27)/41

=14/41

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3 years ago

Which equation is y = 9x2 + 9x – 1 rewritten in vertex form?

Anastasy [175]

Answer:

$y = 9(x +\frac{1}{2}) ^ 2 -\frac{13}{4}$

Step-by-step explanation:

An equation in the vertex form is written as

$y = a (x-h) + k$

Where the point (h, k) is the vertex of the equation.

For an equation in the form $ax ^ 2 + bx + c$ the x coordinate of the vertex is defined as

$x = -\frac{b}{2a}$

In this case we have the equation $y = 9x^2 + 9x - 1$ .

Where

$a = 9\\\\b = 9\\\\c = -1$

Then the x coordinate of the vertex is:

$x = -\frac{9}{2(9)}\\\\x = -\frac{9}{18}\\\\x = -\frac{1}{2}$

The y coordinate of the vertex is replacing the value of $x = -\frac{1}{2}$ in the function

$y = 9 (-0.5) ^ 2 + 9 (-0.5) -1\\\\y = -\frac{13}{4}$

Then the vertex is:

$(-\frac{1}{2}, -\frac{13}{4})$

Therefore The encuacion excrita in the form of vertice is:

$y = a(x +\frac{1}{2}) ^ 2 -\frac{13}{4}$

To find the coefficient a we substitute a point that belongs to the function $y = 9x^2 + 9x - 1$

The point (0, -1) belongs to the function. Thus.

$-1 = a(0 + \frac{1}{2}) ^ 2 -\frac{13}{4}$

$-1 = a(\frac{1}{4}) -\frac{13}{4}\\\\a = \frac{-1 +\frac{13}{4}}{\frac{1}{4}}\\\\a = 9$

<em>Then the written function in the form of vertice is</em>

$y = 9(x +\frac{1}{2}) ^ 2 -\frac{13}{4}$

7 0

3 years ago

Which of the following are solutions to | x-1|= 5x + 2 ? Check all that apply.

agasfer [191]

By definition of absolute value, if $x-1\ge0$ , then $|x-1|=x-1$ , so

$|x-1|=5x+2\implies x-1=5x+2\implies4x=-3\implies x=-\dfrac34$

On the other hand, if $x-1$ , then $|x-1|=-(x-1)=1-x$ , and

$|x-1|=5x+2\implies1-x=5x+2\implies6x=-1\implies x=-\dfrac16$

4 0

2 years ago