I have my maths exam in 20 minutes any tips on how to calm my nerves Its a 2 hour exam :/

1 answer:

3 0

Answer: Try breathing in and out a few times and telling yourself that you will be fine. If this does not work, try chewing gum for a few minutes, the gum gives you something to take your mind off your nerves and also induces a focused headset. Also try to visualize yourself passing the exam with flying colors so you won't worry as much. I hope this helps and your exam goes well! Stay safe.

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HURRY PLZ

AlekseyPX

6^2 + 20.1^2 =? 22.9^2
36 + 404 does NOT equal 524.41
These do NOT form a right triangle.

10.5^2 + 20.8^2 =? 23.3^2
110.25 + 432.64 = 542.89
Therefore, it is a right triangle.

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3 years ago

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Which statement about the graph is true? On a coordinate plane, a line goes through points (0, 2), (2, 3), (4, 4), (6, 5). The g

Tema [17]

The answer is C. The graph does not show a proportional relationship because each point written as a ratio gives a different value.

Hope this helped! If it didn't please tell me what went wrong so I can fix it.

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2 years ago

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Which equation represents the graphed linear function?

FromTheMoon [43]

Answer:

D y=-1/2x+4

Step-by-step explanation:

use the order parir (4,2) x=4

y=-1/2*4+4

y=-2+4

y=2

3 0

2 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

2 years ago

Select all statements that are true about the linear equation y=1/3x+2

VMariaS [17]

Answer:

✨what are the statements.? you didnt label them ✨

Step-by-step explanation:

✨GOD LOVES YOU✨

And welcome to brainly!

3 0

2 years ago