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3 years ago

12

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

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.

Marina86 [1]

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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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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rosijanka [135]

There is a 15/26 chance that it will be a boy

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If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then name the type of quadrilateral

lesya692 [45]

Answer:whats the full questions?

Step-by-step explanation:

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