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2 years ago

Cole noticed that if he takes the opposite of his age and adds 40 he gets the number 28. how old is Cole?

1 answer:

5 0

-x+40=28
-x=-12 (subtract 40 from both sides)

Divide by -1(in both sides)because you cant have a negative age.
COLE is 12

To check plug in the values.

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2. If ¡2x+3 = i, where i = V-1, then which of the

Alexus [3.1K]

I assume you mean the equation

$i^{2x+3} = i$

Since i = √(-1), we have i⁴ = 1 and so i⁵ = i. Then 2x + 3 must be some multiple of 5.

Now just check for which of the given choices that this holds:

x = 16 ⇒ 2x + 3 = 35 (yes)

x = 11 ⇒ 2x + 3 = 25 (yes)

x = 10 ⇒ 2x + 3 = 23 (no)

x = 14 ⇒ 2x + 3 = 31 (no)

6 0

2 years ago

A wire 6 meters long is cut into two pieces. One piece is bent into a square for a frame for a stained glass ornament, while the

zhenek [66]

Answer:

Used wire in circle x = 2.64 m

Used in square L - x = 3.36 m

Total wire used 6 m

Step-by-step explanation:

We have a wire of 6 meters long.

We will cut it a distance x from one end, to get two pieces

x and 6 - x

We are going to use the piece x to get the circle then

So Perimetr of a circle is 2π*r (r is the radius of the circle) then:

x = 2*π*r ⇒ r = x/2*π

And area would be A(c) = π* (x/2*π)² ⇒ A(c) = x²/4π

From 6 - x we will get a square, and as the perimeter is 4 times the side

we have

( 6 - x )/ 4 is the side of the square

And the area is A(s) = [( 6 - x ) /4]²

Total area as function of x is

A(t) = A(c) + A(s)

A(x) = x²/4π + [ ( 6 - x ) / 4 ]²

A(x) = x²/4π + (36 + x² - 12x) /16

A(x) = 1 / 16π [ 4x² + 36π + πx² - 12π x ]

Taking drivatives on both sides of the equation we get:

A´(x) = 1/ 16π [8x +2πx - 12π]

A´(x) = 0 ⇒ 1/ 16π [8x +2πx - 12π] = 0

[8x +2πx - 12π] = 0

8x + 6.28x - 37.68 = 0

14.28x - 37.68 = 0 ⇒ x = 37.68 /14.28

x = 2.64 m length of wire used in the circle

Then the length L for the side of the square is

(6 - x )/4 ⇒ ( 6 - 2.64 )/ 4 ⇒ 3.36 / 4

L = 0.84 m total length of wire used in the square is

3.36 m

And total length of wire used is 6 m

The function is a quadratic function and "a" coefficient is positive then is open upward parabola there is not a maximun

5 0

3 years ago

Estimate the product of 86 and 492

Aleks04 [339]

86 × 492 = about 42,000

7 0

3 years ago

Read 2 more answers

Write an equation of the line that passes through point (2,-4) and has the slope m= 1/2 <br> y =

iren2701 [21]

Answer:

tHe can discuss intelligently the current political affairs. Television is a machine that ought to be used intelligently. It is necessary to think intelligently before performing any action.

Step-by-step explanation:

5 0

2 years ago

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie

ludmilkaskok [199]

Answer:

$\lambda \geq 6.63835$

Step-by-step explanation:

The Poisson Distribution is "a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event".

Let X the random variable that represent the number of chocolate chips in a certain type of cookie. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

$P(X\geq 2)=1-P(X$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}$

$P(X=1)=\frac{e^{-\lambda} \lambda^1}{1!}=\lambda e^{-\lambda}$

And replacing we have this:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]=1-[e^{-\lambda} +\lambda e^{-\lambda}[]$

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)$

And we want this probability that at least of 99%, so we can set upt the following inequality:

$P(X\geq 2)=1-e^{-\lambda}(1+\lambda)\geq 0.99$

And now we can solve for $\lambda$

$0.01 \geq e^{-\lambda}(1+\lambda)$

Applying natural log on both sides we have:

$ln(0.01) \geq ln(e^{-\lambda}+ln(1+\lambda)$

$ln(0.01) \geq -\lambda+ln(1+\lambda)$

$\lambda-ln(1+\lambda)+ln(0.01) \geq 0$

Thats a no linear equation but if we use a numerical method like the Newthon raphson Method or the Jacobi method we find a good point of estimate for the solution.

Using the Newthon Raphson method, we apply this formula:

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$

Where :

$f(x_n)=\lambda -ln(1+\lambda)+ln(0.01)$

$f'(x_n)=1-\frac{1}{1+\lambda}$

Iterating as shown on the figure attached we find a final solution given by:

$\lambda \geq 6.63835$

4 0

3 years ago