Could someone please simplify for me the square root of 4312?? I would really appreciate it.

Which could be the graph of f(x) = |x -h| + k if h and k are both positive?

Nesterboy [21]

Step-by-step explanation:

The graph of the function y=|x| is as shown in the attached diagram. The only possible translation of this graph right and up is option A. (In option B graph is translated right and down, in option C- left and up and in option D - left and down). Answer: correct choice is A.

3 0

3 years ago

How many different 3-digit PIN codes are there that only include the digits 7, 4, 3 and 2?

KengaRu [80]

Answer:

64

Step-by-step explanation:

We can use the product rule of counting to answer this:

There are 4 possibilities for the 1st digit
There are 4 possibilities for the 2nd digit
There are 4 possibilities for the 3rd digit
So we use the product rule of counting to give us:

4×4×4=64

6 0

2 years ago

If y=e5t is a solution to the differential equation

a_sh-v [17]

Answer:

k = 30, $y(t) = C_1e^{5t}+C_2e^{6t}$

Step-by-step explanation:

Since $y=e^{5t}$ is a solution, then it must satisfy the differential equation. So, we calculate the derivatives and replace the value in the equation. We have that

$\frac{d^2y}{dt^2} = 25 e^{5t},\frac{dy}{dt} = 5e^{5t}$

Then, replacing the derivatives in the equation we have:

$25e^{5t}-11(5)e^{5t}+ke^{5t}=0 e^{5t}(25-55+k) =0$

Since $e^{5t}$ is a positive function, we have that

$25-55+k = 0 \rightarrow k = 30$ .

Now, consider a general solution $y(t) = Ae^{rt}, A \in \mathbb{R}$ , then, by calculating the derivatives and replacing them in the equation, we get

$Ae^{rt}(r^2-11r+30)=0$

We already know that r=5 is a solution of the equation, then we can divide the polynomial by the factor (r-5) to the get the other solution. If we do so, we get that (r-6)=0. So the other solution is r=6.

Therefore, the general solution is

$y(t) = C_1e^{5t}+C_2e^{6t}$

8 0

3 years ago

A company manager for a tire manufacturer is in charge of making sure there is the least amount of defective tires. If the tire

Anestetic [448]

Answer:

0.347% of the total tires will be rejected as underweight.

Step-by-step explanation:

For a standard normal distribution, (with mean 0 and standard deviation 1), the lower and upper quartiles are located at -0.67448 and +0.67448 respectively. Thus the interquartile range (IQR) is 1.34896.

And the manager decides to reject a tire as underweight if it falls more than 1.5 interquartile ranges below the lower quartile of the specified shipment of tires.

1.5 of the Interquartile range = 1.5 × 1.34896 = 2.02344

1.5 of the interquartile range below the lower quartile = (lower quartile) - (1.5 of Interquartile range) = -0.67448 - 2.02344 = -2.69792

The proportion of tires that will fall 1.5 of the interquartile range below the lower quartile = P(x < -2.69792) ≈ P(x < -2.70)

Using data from the normal distribution table

P(x < -2.70) = 0.00347 = 0.347% of the total tires will be rejected as underweight

Hope this Helps!!!

6 0

3 years ago

The greatest common factor of 3m2n + 12mn2 is

Karolina [17]

$3m^2n + 12mn^2=3mn\cdot(m+4n)$