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2 years ago

In how many ways can a teacher arrange 5 students in the front row of a classroom with a total of 20 students?

Mathematics

1 answer:

makkiz [27]2 years ago

7 0

C, 60,480 i took thiis before it easy as

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What are the powers of 4 through 1000

ahrayia [7]

Answer:

I'm pretty sure that they are 4, 16, 64, and 256.

Step-by-step explanation:

4 0

2 years ago

In a circle with a radius of 5cm and is centered at point O, angle AOB intercepts arc AB. Arc AB has a length of 10cm. What is t

Elenna [48]

circumference of circle = 2*pi*r given r = 5 so circumference = 31.42 now we know 360 degree is uniformly distributed on this circumference so x/360=10/31.42 so x= 114.65 so option C.) 114.65 is the answer

4 0

2 years ago

Y=x+ 5 y = 6 HELP!! Solve this system of equation algerbraically.

kati45 [8]

Answer:

x = 1

Step-by-step explanation:

plug y = 6 on to the equation y =x + 5

=> 6 = x +5

=> x = 1

5 0

2 years ago

Read 2 more answers

Write the quadratic equation y= (x)2-6x-7 in vertex form

lakkis [162]

Answer:

y = (x-3)^2 -16

Step-by-step explanation:

y = x^2 -6x-7

We need to complete the square

b= -6

b/2 = -6/2 = -3

(b/2) ^2 = (-3)^2 =9

Add 9 then subtract 9 to balance the equation

y = x^2 -6x +9 -9 -7

y = (x^2 -6x+9) -16

The terms in parentheses is (x-b/2)^2

y = (x-3)^2 -16

This is in vertex form

The vertex is (3,-16)

4 0

2 years ago

Read 2 more answers

In a murder investigation, the temperature of a corpse was 32.5 degrees C at 1:30 pm and 30.3 degrees C an hour later. Normal bo

svp [43]

Let $T(t)$ be the temperature of the body $t$ hours after 1:30 PM. Then $T(0) = 32.5$ and $T(1) = 30.3$ .

Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T-20)$ . Now let $y = T - 20$ , so
$y(0) = T(0) - 20 = 32.5 - 20 = 12.5$ , so $y$ is a solution to the initial value problem $dy/dt = ky$ with $y(0) = 12.5$ .
By separating and integrating, we have $y(t) = y(0)e^{kt} = 12.5e^{kt}$ .

$y(1) = 30.3 - 20\ \Rightarrow\ 12.5e^{k(1)} = e^{k} = \frac{10.3}{12.5}\ \Rightarrow \\
k = \ln \frac{10.3}{12.5}
$

$y(t) = 37 - 20\ \Rightarrow\ 12.5e^{kt} = 17\ \Rightarrow\ e^{kt} = \frac{10.3}{12.5} \ \Rightarrow \\ \\
kt = \ln \frac{10.3}{12.5}\ \Rightarrow\ t = \left( \ln \frac{17}{12.5} \right)/ \frac{10.3}{12.5} \approx -1.588 \mathrm{\ h}$

≈ 95 minutes. Thus the murder took place about 95 minutes before 1:30 PM, or 11:55 AM.

6 0

2 years ago