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ddd [48]
3 years ago
12

If n^3=x and n^4=20x , where n>0, what is the value of x?

Mathematics
1 answer:
Natali5045456 [20]3 years ago
6 0

Step-by-step explanation:

∠2 and ∠7 are alternate exterior angles so they are equal.

So you can create the equation

40x-48 = 20x+52   Subtract 20x from both sides. Add 48 to both sides.

20x = 100

x = 5

if you need  m∠2 Substtute 5 for x in 20x+52  

20(5) + 52 = 100 +52   m∠2= 152°

m∠7  40(5) - 48 =  200-48  m∠7= 152°

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Question number 8. Prove that 3 + √5 is an irrational number.​
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Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

\\  \sf \: 3 +  \sqrt{5}  =  \frac{a}{b}   \\  \\   \qquad \: \tiny \sf{(where \:  \: a \:  \: and \:  \: b \:  \: are \:  \: integers \:  \: and \:  \: b \:  \neq \: 0)} \\

\\  \sf \:  \sqrt{5}  =  \frac{a}{b}  - 3 =  \frac{a - 3b}{b}  \\

Since, a, b and 3 are integers. So,

\\ \sf \:  \frac{p - 3b}{b}  \\  \\  \qquad \tiny \sf{ \: (is \:  \: a \:  \: rational \:  \: number \:) } \\

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

\\

Hence, 3 + √5 is an irrational number.

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3 years ago
PLS HELP, AM BEING TIMED, WILL give brainliest and it's a test​
barxatty [35]
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5 0
3 years ago
On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)
xz_007 [3.2K]

Answer:

P = 10 + 2\sqrt{53} units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3),  X (2, 3),

Y (4, -4),  Z (-3, -2)

Required

Determine perimeter of the kite

First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

For WX:

(x_1, y_1)\ (x_2,y_2) = (-3, 3),\ (2, 3)

WX = \sqrt{(-3 - 2)^2 + (3 - 3)^2}

WX = \sqrt{(-5)^2 + (0)^2}

WX = \sqrt{25}

WX = 5

For XY:

(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)

XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}

XY = \sqrt{-2^2 + (3 +4)^2}

XY = \sqrt{-2^2 + 7^2}

XY = \sqrt{4 + 49}

XY = \sqrt{53}

For YZ:

(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)

YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}

YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}

YZ = \sqrt{7^2 + (-2)^2}

YZ = \sqrt{49 + 4}

YZ = \sqrt{53}

For ZW:

(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)

ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}

ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}

ZW = \sqrt{0^2 + (-5)^2}

ZW = \sqrt{0 + 25}

ZW = \sqrt{25}

ZW = 5

The Perimeter (P) is as follows:

P = WX + XY + YZ + ZW

P = 5 + \sqrt{53} + \sqrt{53} + 5

P = 5 + 5 + \sqrt{53} + \sqrt{53}

P = 10 + 2\sqrt{53} units

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Answer:whats the full questions?

Step-by-step explanation:

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