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3 years ago

Is x=-7 an equation that describes a function

2 answers:

5 0

Xy + 7 = x + y is not a linear equation because the term xy has degree 2. x + 3y2 = 6 is not a linearequation because the term 3y2 has degree 2. ... x = (constant). is a linear equation but does notdescribe a function

Oksana_A [137]3 years ago

4 0

No it is not. unless it doesnt pass threw another point

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Awnser this for braainliest pls

Ilya [14]

Answer:

4.5 so 5

Step-by-step explanation:

36 divided by 8= 4.5

5 0

3 years ago

Read 2 more answers

COMPLETE The equation X-9=0 has how many real solution(s).

Maru [420]

Answer:

one solution

Step-by-step explanation:

X-9=0

Add 9 to each side

X-9+9=0+9

x = 9

There is one solution

5 0

3 years ago

Isaac has a bag of n peanuts. He shares the peanuts with 5 of his friends. Each person gets at least 18 peanuts. Write an inequa

Setler [38]

Answer:

Total number of peanuts =n

He shared with 5 people = 6 people

Each person had at least 18

n/6>=18

Step-by-step explanation:

When question like this is given , the English is very important. He shares with his friends is different from he shares among his friends. Then also, at least means greater than or equal to.

4 0

3 years ago

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Two students are reading a novel. Ashley reads 10 pages per day. Carly reads 8 pages per day, but she starts early and is alread

Alecsey [184]

Is there more options because somethings missing

4 0

3 years ago

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