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xeze [42]
3 years ago
12

on every three hamburgers that McDonald's makes, they use 9pickles. how many pickles would they use on 100 hamburgers

Mathematics
1 answer:
Feliz [49]3 years ago
6 0
The answer is easy 900
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Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)



\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\

4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}
3 0
3 years ago
Solve the linear equation for x. –4.8(6.3x – 4.18) = –58.56 x =
liq [111]

Answer:

X=2.6

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
How do I get the answer y= 2x^2+9x-5
guapka [62]

Answer:

Quadratic Formula

so

x = -5

and

x = 0.5

Step-by-step explanation:

Whenever you see a problem in this form, which you will see a lot of, you can try to factor it or use the "least squares" method or what have you, but those won't always work, unfortunately.

Fortunately, the quadratic formula will never fail you with quadratic expressions.

This is the Quadratic Formula

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

a is the the number on the variable with the exponent ^2

b is the number on the variable with no exponent

c is the third number

a and b cannot be equal to 0; c can be

Since we're looking for a number with an equation that has a square root in it, we're going to get two answers. These two answers come from the radical being separately added AND subtracted from the radical. It's basically two problems.

Plugging  in our numbers to this equation gives us x values of -5 and 0.5. This will always work with polynomials with factors of ^2 in them.

If you have a TI-84 calculator or newer, there's a tool on it that will factor polynomials like this one for you just by giving it the numbers.

7 0
3 years ago
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