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2 years ago

Find the vale of x help please

Mathematics

2 answers:

dezoksy [38]2 years ago

4 0

So u got opposite and hypotenuse right? So that means u need to use O/H so that means you need to use sine. Therefore this is trigonometry and 4/sin 45 is ur answer mate alright? Tossed

krek1111 [17]2 years ago

4 0

Answer:

the answer is 4 with a square root of 2

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Given that a random variable X has a geometric distribution that X- Geometry 0.67. Find the mean

yarga [219]

Answer:

Hello,

0,4925...

Step-by-step explanation:

A geometric random variable X with parameter p has probability function:

$f(x)=p*(1-p)^x\ where\ x=0,1,2,3....\\\\mean=\dfrac{1-p}{p} =\dfrac{1-0.67}{0.67} =0.4925...$

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2 years ago

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likoan [24]

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A sign chart is a chart which help in solving polynomial inequalities.

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2 years ago

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1 and 2 pls help me with these math problems.........

jeka57 [31]

1. y=4x

2. (I am not able to graph it, but here are some points on the line.)
(0,3) (3,6) (5,8) those three should get you to graph a line. When you have these types of problems, make a table. plug in random numbers, that are reasonable, usually 1-6, and plug them into the equation and enter your new y values, then graph the ordered pairs and connect them.

6 0

3 years ago

Ellus If f(x) = x2, and g(x) = x - 1, then f(g(x)) = [ ? ]x2 +[ ]x+]

Nostrana [21]

Answer:

x^2-2x+1

Step-by-step explanation:

f(x)=x^2

g(x)=x-1

f(g(x))=f(x-1)=(x-1)^2=x^2-2x+1

8 0

2 years ago

HELp meh with this question it very hard

Dovator [93]

Answer:

AB = 7 cm

Step-by-step explanation:

Use the law of cosines to find the side AB:

$AB = \sqrt{(x + 3)^2+x^2-2x(x+3)cos (60)} =$
$\sqrt{x^2+6x+9+x^2-x^2-3x} = \sqrt{x^2+3x+9}$

Use the Heron's area formula next:

$A = \sqrt{s(s - a)(s-b)(s-c)}$ , where s- semi perimeter

s = 1/2[x + x + 3 + $\sqrt{x^2+3x+9}$ ) = 1/2 (2x + 3 + $\sqrt{x^2+3x+9}$ )
s - a = 1/2 (2x + 3 + $\sqrt{x^2+3x+9}$ - 2x - 6) = 1/2 ( $\sqrt{x^2+3x+9 }$ - 3)
s - b = 1/2 (2x + 3 + $\sqrt{x^2+3x+9}$ - 2x) = 1/2 ( $\sqrt{x^2+3x+9}$ + 3)
s - c = 1/2 (2x + 3 + $\sqrt{x^2+3x+9}$ - 2 $\sqrt{x^2+3x+9}$ ) = 1/2 (2x + 3 - $\sqrt{x^2+3x+9}$ )

Now

(s - a)(s - b) = 1/4 [(x²+3x+9) - 9] = 1/4 (x² + 3x)
s(s - c) = 1/4 [(2x + 3)² - (x² + 3x + 9)] = 1/4 (3x²+ 9x) = 3/4(x² + 3x)

Next

A² = 3/16(x² + 3x)(x² + 3x)
300 = 3/16(x² + 3x)²
1600 = (x² + 3x)²
x² + 3x = 40

Substitute this into the first equation:

$AB = \sqrt{40 + 9} = 7 cm$

4 0

3 years ago