The metric ruler is divided into

What have I done wrong and where please help me

lys-0071 [83]

12x - 18 - 4x -2
12x - 4x = 8x (yours 12x - 4x = 8)
-18 - 2 = -20 (yours -18 -1 = 17)

answer:

8x - 20

6 0

3 years ago

Please help! I’ve tried every site and nothing has helped<br><br><br>The answer is 11.8

melamori03 [73]

Answer:

11.8%

Step-by-step explanation:

Here in this question, we want to find the probability of no success in the binomial experiment for 6 trials.

Let p = probability of success = 30% = 30/100 = 0.3

q = probability of failure = 1-p = 1-0.3 = 0.7

Now to calculate the probability, we shall need to use the Bernoulli approximation of the binomial theorem.

That would be;

P(X = 0) = 6C0 p^0 q^6

6C0 is pronounced six combination zero

= 6 * 0.3^0 * 0.7^6 = 1 * 1 * 0.117649 = 0.117649

This is approximately 0.1176

If we convert this to percentage we have 11.76%

But we want our answer rounded to the nearest tenth of a percent and that is 11.8%

8 0

3 years ago

One fourth dived by one fourth

viva [34]

1; (1/4)/(1/4), all you do is flip the fraction to make the reciprocal and multiply, so (4/1)x(1/4)=1

8 0

3 years ago

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

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<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

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<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0

2 years ago

Cube root of b to the 12 power

kvv77 [185]

$\sqrt[3]{b^{12}}=\sqrt[3]{(b^4)^3}=b^4$

6 0

3 years ago