Please help: Triangle ABC is similar to triangle PQR where

A company had 41 employees in order 980 uniforms for them if they want to give each employee same number of uniforms how many mo

Tema [17]

Answer:

They would have to order 4 more uniforms in order to distribute an equal amount to each employee

Step-by-step explanation:

First we have to calculate the number of maximum uniforms that can be given to each employee equally

For this we simply divide the number of uniforms by the number of employees and look only at the whole number

980/41 = 23.92 = 23

we don't round we just take the decimals

now we multiply the number of maximum uniforms that we can give each one by the number of employees

23 * 41 = 943

to the 980 uniforms we subtract the 943

980 - 943 = 37

Calculate how much is left to 37 to reach 41

41 - 37 = 4

This means that they would have to order 4 more uniforms in order to distribute an equal amount to each employee

3 0

3 years ago

The smallest multiple of 50

trasher [3.6K]

Answer:

1

Step-by-step explanation:

One is the smallest multiple of 50. We can find this by finding all the multiples of 50. The multiples of 50 are: 1, 2, 5, 10, 25, and 50.

We can see that one is the smallest multiple because it is less than 2, 5, 10, 25 and 50. So, that means the answer to the question is 1.

Hope this helps.

6 0

2 years ago

Read 2 more answers

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre

Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = $\frac{d}{dx}$ [ $x^{4}ln(x)$ ]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = $4x^{3}ln(x) + x_{4}.\frac{1}{x}$

f'(x) = $4x^{3}ln(x) + x^{3}$

f'(x) = $x^{3}[4ln(x) + 1]$

Now, find the critical points: f'(x) = 0

$x^{3}[4ln(x) + 1]$ = 0

$x^{3} = 0$

x = 0

and

$4ln(x) + 1 = 0$

$ln(x) = \frac{-1}{4}$

x = $e^{\frac{-1}{4} }$

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval x-value f'(x) result

0<x<0.78 0.5 f'(0.5) = -0.22 decreasing

x>0.78 1 f'(1) = 1 increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = $x^{4}ln(x)$

f(0.78) = $0.78^{4}ln(0.78)$