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3 years ago

12

Test your PEMDAS 6+8/2(2+2) This one is tricky

1 answer:

Ivanshal [37]3 years ago

7 0

Do parenthesis first (4)
Then, do the division 8/2 = 4
Then multiply 4 and 4 which will be 16
Add 6 to 16 and you get 22

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What os the best definition of a triangle

pav-90 [236]

D. Becaus it is the only one that mentions the characters of a triangle

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3 years ago

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Please help!!! i will give brainliest

belka [17]

Susan has 50 so the difference is 5 sense sheila had 45

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3 years ago

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I need to know this quick! 10 points is Max sorry

Stolb23 [73]

Answer:

A

Step-by-step explanation:

$\frac{5}{6}\cdot \frac{1}{2}=\frac{5}{12}$

Multiply top and bottom, on the top we get 5*1 which is 5 and on the bottom we get 6*2 which is 12.

7 0

3 years ago

Will mark brainliest!!!plz helppp

muminat

Answer:

(5,-6)

Step-by-step explanation:

ONE WAY:

If $f(x)=x^2-6x+3$ , then $f(x-2)=(x-2)^2-6(x-2)+3$ .

Let's simplify that.

Distribute with $-6(x-2)$ :

$f(x-2)=(x-2)^2-6x+12+3$

Combine the end like terms $12+3$ :

$f(x-2)=(x-2)^2-6x+15$

Use $(x-b)^2=x^2-2bx+b^2$ identity for $(x-2)^2$ :

$f(x-2)=x^2-4x+4-6x+15$

Combine like terms $-4x-6x$ and $4+15$ :

$f(x-2)=x^2-10x+19$

We are given $g(x)=f(x-2)$ .

So we have that $g(x)=x^2-10x+19$ .

The vertex happens at $x=\frac{-b}{2a}$ .

Compare $x^2-10x+19$ to $ax^2+bx+c$ to determine $a,b,\text{ and } c$ .

$a=1$

$b=-10$

$c=19$

Let's plug it in.

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

$\frac{-(-10)}{2(1)}$

$\frac{10}{2}$

$5$

So the $x-$ coordinate is 5.

Let's find the corresponding $y-$ coordinate by evaluating our expression named $g$ at $x=5$ :

$5^2-10(5)+19$

$25-50+19$

$-25+19$

$-6$

So the ordered pair of the vertex is (5,-6).

ANOTHER WAY:

The vertex form of a quadratic is $a(x-h)^2+k$ where the vertex is $(h,k)$ .

Let's put $f$ into this form.

We are given $f(x)=x^2-6x+3$ .

We will need to complete the square.

I like to use the identity $x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2$ .

So If you add something in, you will have to take it out (and vice versa).

$x^2-6x+3$

$x^2-6x+(\frac{6}{2})^2+3-(\frac{6}{2})^2$

$(x+\frac{-6}{2})^2+3-3^2$

$(x+-3)^2+3-9$

$(x-3)^2+-6$

So we have in vertex form $f$ is:

$f(x)=(x-3)^2+-6$ .

The vertex is (3,-6).

So if we are dealing with the function $g(x)=f(x-2)$ .

This means we are going to move the vertex of $f$ right 2 units to figure out the vertex of $g$ which puts us at (3+2,-6)=(5,-6).

The $y-$ coordinate was not effected here because we were only moving horizontally not up/down.

3 0

3 years ago

Fast pleaseeeeeeee.... Find x

r-ruslan [8.4K]

X = 360 - (158 + 148 + 32)
x = 360 - 338
x = 22

3 0

3 years ago