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2 years ago

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5/7 + 3/8?

1 answer:

Rudiy272 years ago

7 0

I don't know if anyone know say him answer

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Given that p= x-y <br> Find p when :<br> X=-12. And y=-19<br><br> What is p?

Mazyrski [523]

Answer:

p=7

Step-by-step explanation:

p=-12-(-19)

p=-12+19

p=7

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2 years ago

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A random number generator app on your phone generates two numbers between 0 and 8. What is the probability that the first number

aev [14]

Answer:

1/27

Step-by-step explanation:

There are 9 numbers between 0 and 8. 3 of them are less than 3 (0, 1, and 2), and 1 of them is 6.

The probability is therefore (3/9) (1/9) = 1/27.

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2 years ago

What is the sum of 6 ×10^3 and3×10^2?

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Step-by-step explanation:

6000+300

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2 years ago

Which is closet to 1,0, or 1 half

Alekssandra [29.7K]

36/40 would be close to 1

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3 years ago

Read 2 more answers

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.

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2 years ago