Ask question

Ask question

All categories

3 years ago

5

If x is the product of the integers from 1 to 150, inclusive, and 5^y is a factor of x, what is the greatest possible value of y

?
A) 30
B) 34
C) 36
D) 37
E) 39

1 answer:

Maslowich3 years ago

5 0

ANSWER: A) 30

5×30=150 which is the highest number you can go to

You might be interested in

To the nearest tenth, what is the area of the shaded segment when AG = 6 ft?

valentinak56 [21]

A. 10.3 assuming that 113.1 is given

7 0

3 years ago

Read 2 more answers

What is the percent change if the student population of the school goes from 380 students to 335 students? Round to the nearest

lys-0071 [83]

Answer:

12% decrease

Step-by-step explanation:

I'm srry if I'm wrong but I'm 90-ish% its correct

6 0

3 years ago

Read 2 more answers

Helppppppppppp. Pleaseeeeeeeeeeeeeeeeeeeeeeeeeeee

olga55 [171]

Answer:

7

Step-by-step explanation:

form a proportion

12/2=42/x

cross multiply

12x=84

x=7

6 0

2 years ago

Read 2 more answers

Two functions are shown below.

Angelina_Jolie [31]

Answer: 3

Step-by-step explanation:

Given

$f(x)=\dfrac{1}{3}3^x$

$g(x)=2x+4$

For $f(x)$ to be true, the domain is given by observing the graph of two i.e. $x\in (-1.981,3.117)$ . In this range $g(x)$ is greater than $f(x)$ .

From the above domain largest integer that can satisfy the condition is 3.

3 0

3 years ago

Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En

ArbitrLikvidat [17]

Answer:

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ , c) $x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$ , $y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$ .

Step-by-step explanation:

The equation of the circle is:

$x^{2} + (y-1)^{2} = 16$

After some algebraic and trigonometric handling:

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1$

$\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t$

Where:

$\frac{x}{4} = \cos t$

$\frac{y-1}{4} = \sin t$

Finally,

$x = 4\cdot \cos t$

$y = 1 + 4\cdot \sin t$

a) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

b) $x = 4\cdot \cos t$ , $y = 1 + 4\cdot \sin t$ .

c) $x = 4\cdot \cos t''$ , $y = 1 + 4\cdot \sin t''$

Where:

$4\cdot \cos t' = 0$

$1 + 4\cdot \sin t' = 5$

The solution is $t' = \frac{\pi}{2}$

The parametric equations are:

$x = 4\cdot \cos \left(t+\frac{\pi}{2} \right)$

$y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)$

7 0

3 years ago