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Alecsey [184]
2 years ago
8

Kelley ate 3/5 of her brownies on Monday and 6/8 of her brownies Tuesday. If Kelley originally had 2 brownies, what fraction bro

wnie is left over? Simplifty your answer
Mathematics
1 answer:
Sergeeva-Olga [200]2 years ago
8 0
<h2>Greetings!</h2>

Answer:

\frac{13}{20}

Step-by-step explanation:

First, we need to add the two fractions together. To do this we need to make the denominator a factor of both 8 and 5. Easy way to do this is to multiply each fraction by the denominator of the other fraction:

\frac{3}{5} * 8 = \frac{24}{40}

\frac{6}{8} * 5 = \frac{30}{40}

\frac{24}{40} + \frac{30}{40} = \frac{54}{40}

We can also show 2 brownies as \frac{2}{1} or \frac{80}{40}

Now we can subtract the two:

\frac{80}{40} -  \frac{54}{40} = \frac{26}{40}

Simplify the fraction down by dividing by two:

\frac{13}{20}

Which is the fraction of brownie left!


<h2>Hope this helps!</h2>
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Step-by-step explanation:

Rate is simplest form, so 3 napkins for 5 cents.

Unit rate is for 1 x value, so 1 napkin for 0,6 cents.

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2 years ago
How many solutions does this
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Erm- can’t answer this. We need the graph
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3 years ago
Topic: The Quadratic Formula
Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

x= \frac{-b +- \sqrt{b^2-4ac} }{2a}

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² =  x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\=  -4 or 1

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

7 0
3 years ago
Simplify (3a-2b)²-2(3a-2b)(a+2b)+(a+2)²​
Mnenie [13.5K]

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the 2 is squareing 4a and 12b

Step-by-step explanation:

expand move the parentheses

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6 0
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