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3 years ago

8

A spotlight at a dance club flashes every 25 seconds. Another spotlight flashes every 30 seconds. Both lights just flashed . Aft

er how many minutes will both lights flash at the same time again?

2 answers:

bazaltina [42]3 years ago

8 0

find the least common multiple of 25 and 30, which is 150

so they will flash at the same time every 150 seconds which equals 2.5 minutes

Vsevolod [243]3 years ago

7 0

150 seconds for both the lights to flash.

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Proof by Contradiction : Show that √ 2 is irrational.

myrzilka [38]

Answer:

$\sqrt2$ is irrational

Step-by-step explanation:

Let us assume that $\sqrt2$ is rational. Thus, it can be expressed in the form of fraction $\frac{x}{y}$ , where x and y are co-prime to each other.

$\sqrt2$ = $\frac{x}{y}$

Squaring both sides,

$2 = \frac{x^2}{y^2}$

Now, it is clear that x is an even number. So, let us substitute x = 2u

Thus,

$2 = \frac{(2u)^2}{y^2}\\y^2 = 2u^2$

Thus, $y^2$ is even, which follows the fact that y is also an even number. But this is a contradiction as x and y have a common factor that is 2 but we assumed that the fraction $\frac{x}{y}$ was in lowest form.

Hence, $\sqrt2$ is not a rational number. But $\sqrt2$ is a an irrational number.

5 0

3 years ago

When Lorenzo finished his exam last week, he thought the test was over. But the instructor put z-scores on each student's paper

Anarel [89]

Answer:

Lorenzo's score on exam was 75.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 61

Standard Deviation, σ = 8

We are given that the distribution of test score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

z-score = 1.75

We have to find the value of x.

Putting values, we get

$1.75 = \dfrac{x - 61}{8}\\\\x = 1.75(8) + 61\\x = 75$

Thus, Lorenzo's score on exam was 75.

5 0

3 years ago

What is the value of d?

ANTONII [103]

The value of 'd' would be 80.

7 0

3 years ago

ethan allens soccer team is having a car wash the team spent $44 on supplies they earned $300 in total the teams profit is the a

grigory [225]

Answer:

the profit is $256

Step-by-step explanation:

300 - 44 = 256

6 0

2 years ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

2 years ago