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3 years ago

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one day, bill at the candy shoo sold 210 bottles of cherry soda and grape soda for a total of 230.30. If the cherry soda costs 1

.15 and the grape soda costs $.99, how many of each kind where sold

1 answer:

oksian1 [2.3K]3 years ago

5 0

Mmmmmmmm””6098876544

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27.64x3 estimate product

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Step-by-step explanation

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Please write step by step of how you got the answer. Thankyou

stealth61 [152]

Answer:

The solution in the attached figure

Step-by-step explanation:

step 1

Place the numbers -4, 4, 0, and -3 in the left side

Adds the numbers

$-4+4+0-3=-3$ ----> is correct

step 2

Place the numbers 3, -1, and -2 in the right side

Adds the numbers

$-3+3-1-2=-3$ ----> is correct

step 3

Place the numbers 1 and 2 in the base side

Adds the numbers

$-4+1+2-2=-3$ ----> is correct

The solution in the attached figure

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3 years ago

(3+62) x (25-17) + 3*2 <br> 2 is an exponent btw

MAXImum [283]

Answer:

529

Step-by-step explanation:

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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.

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2 years ago