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lyudmila [28]
3 years ago
9

A floor plan is given for the first floor of a new house. One inch represents 6 feet. If a small closet is one-fourth in by one-

third in on the floor plan, what are the actual dimensions?
Mathematics
2 answers:
Artist 52 [7]3 years ago
7 0
3/2 and 2 feet yo answer!!!
melomori [17]3 years ago
3 0
3/2 feet and 2 feet is the answer respectively
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The coordinates of an ordered pair have opposite signs. In which quadrant(s) must the ordered pair lie? Explain.
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Alright...so the coordinates of an ordered pair have opposite signs [one sign is positive while the other is negative] so we could have an example of (-x,+y) or (+x,-y) ...that means out of the 4 quadrants these points could be in the 2nd quadrant or the 4th quadrant or corners of the graph

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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}}
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blagie [28]

crosses x-axis at (2, 0 ) and y-axis at (0, - 4 )

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x = 0 : y = 0 - 4 = - 4 ⇒ (0, - 4 )

y = 0 : 2x - 4 = 0 ⇒ 2x = 4 ⇒ x = 2 ⇒ (2, 0 )


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Can anyone please help me with these sums? I need it fast......Please give correct answers only. I will give brainliest.
soldier1979 [14.2K]

Answer:

rise/run= -20/5= -4/1=-4 slope

or you sound use the slope formula (y2-y1)/(x2-x1)

(-9-11)/(1+4)

-20/5

-4

Step-by-step explanation:

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3 liters of soda
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