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3 years ago

3х + 9 = 10x + 9 - 7х Solution? CAN I PLEASEE GET HELPPP ASAPPP!

Mathematics

1 answer:

vampirchik [111]3 years ago

4 0

The answer is equal to each other at my school we call it no solution

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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

Linda is one of 8 teachers in the Mathematics department. In how many different ways can three of the teachers be chosen so that

Alexxx [7]

Answer:

21 ways

Step-by-step explanation:

a, b, c, d, e, f, g, linda

1 2 linda

a__ b__ __

a__c __ __

a__d __ __

a__ e__ __

a__ f__ __

a__ g__ __

b__c __ __

b__d __ __

b__ e__ __

b__ f__ __

b__ g__ __

c__d __ __

c__ e__ __

c__ f__ __

c__ g__ __

d__ e__ __

d__ f__ __

d__ g__ __

e__ f__ __

e__ g__ __

f__ g__ __

count them

in total

there are

21 triples

4 0

3 years ago

Give two examples of line segments that will make a triangle ?

77julia77 [94]

Answer:

If the length of any one side is greater than the sum of the length of the other two, the line segments cannot be used to create a triangle. It is possible to create a triangle using 3 line segments if the sum of the lengths of any two line segments is greater than the length of the third.

7 0

3 years ago

Use the following statements to find a compound statement for (p∨q)∧rp∨q)∧r p: 5<-3 q: All vertical angles are congruent. r:

guajiro [1.7K]

Given the compound statement <span>(p∨q)∧r
where: p: 5 < -3
q : All vertical angles are congruent.
r: 4x = 36, then x = 9.

Recall the in logic, '</span>∨' symbolises "or" while '∧' symbolises "and".

Therefore, the compound statement <span>(p∨q)∧r can be written as follows:
Either 5 < -3 or all vertical angles are congruent, and if 4x = 36, then x = 9.
</span>

4 0

4 years ago

Round your answer to the nearest hundredth AC=

KatRina [158]

Answer:

AC ≈ 2.96

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos65° = $\frac{adjacent}{hypotenuse}$ = $\frac{AC}{AB}$ = $\frac{AC}{7}$