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3 years ago

Find the value of x.

Mathematics

2 answers:

nikdorinn [45]3 years ago

8 0

Answer:

X=11

Step-by-step explanation:

(10x+30)=140

-30 both sides

10x=110

Divide 10 both sides

X=11

KATRIN_1 [288]3 years ago

7 0

x=11

------------------------------------------------------------------------------------------------------

Since 140 and 10x+30 are vertical angles, they are equal.

Equation: 140=10x+30

Subtract 30 from both sides of the equation.

140=10x

Divide both sides of the equation by 10.

You get x=10

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Hope this helped!!

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A school play sells adult and child tickets. On Friday they sold 55 child tickets and 20 adult tickets for $200. On Saturday the

Anarel [89]

Answer:

Step-by-step explanation:

Let, price of individual child ticket is $x, and price of individual adult ticket is $y.

Now, on Friday we can write an equation is 55x+20y=200....(1).

and on Saturday we can write an equation is 72x+36y=306.....(2).

so, Equation 1: 55x+20y=200

Equation 2: 72x+36y=306

now, we are going to solve both equations.

after solving you will find x=2 and y=4.5

Hence, we can say that the cost of child ticket is $2, and the cost of adult ticket is $4.50.

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3 years ago

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

Finally
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\boxed{\boxed{x=1}}$

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