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2 years ago

Solve the system of equations y=x^2-3x+4 x+y=4 show all work please!

Mathematics

1 answer:

Debora [2.8K]2 years ago

6 0

Answer:

(0,4)

(2,2)

Step-by-step explanation:

y=x^2-3x+4

x+y=4

the second one can be rewritten as y=4-x

so the system becomes

y=x^2-3x+4

y=4-x

therefore 4-x = x^2-3x+4

that is x^2-2x=0

we can write it as x(x-2)=0 a product is = 0 when one of its factors = 0

so the solutions for x are: x=0 and x-2= 0 that means x=2

to get y you simply use 0 and 2 as value of x for any of the two equations

and you get y=4 and y=2 respectively

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NARA [144]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Paolo is using his grandmother’s cookie recipe. He always doubles the amount of chocolate chips and oats. The recipe calls for 2

nalin [4]

<u>Answer:</u>

y = 94.5 cups

<u>Step-by-step explanation:</u>

Let's say:

y = oats
x = chocolate chips = 212

=> Full Recipe = 212 + y

=> 2(Full Recipe) = 2(212 + y) = 613 cups

=> 2(Full Recipe) = 424 + 2y = 613 cups

=> 2y = 613 - 424

=> 2y = 189

=> y = 94.5

Therefore, the required oats to do double the recipe is 94.5 cups.

Hoped this helped.

$GeniusUser$

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2 years ago

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Answer:

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2 years ago

Identify a pattern and fine the next number in the pattern.

aleksklad [387]

The answer is C cuz you're dividing by 3

3 0

2 years ago

What’s is the answer

kvasek [131]

When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

$(x^{2} )^{3} = x^6$

$(x^{3} )^5=x^{15}$

When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

$(x^{2} )(x^{3})=x^{5}$

$(x^{1} )(x^{2})=x^{3}$

$(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}$

You multiply 3 into each exponent in the numerator and the denominator

$\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}$

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive.

$\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}$

When you have something like this:

$\frac{x^{2}}{x^5}$

You subtract the exponents together, so:

$\frac{x^2}{x^5} = x^{2-5} = x^{-3} = \frac{1}{x^3}$

Your answer is the second option

3 0

2 years ago