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3 years ago

Where does the minute hand point at ten past seven

Mathematics

2 answers:

o-na [289]3 years ago

6 0

Answer:

The minute hand would point to the 2.

Step-by-step explanation:

You count the numbers on the clock by 5's. So, when the minute hand is on 1, it's 5 after, when it hits 2, it will be 10 after.

Ronch [10]3 years ago

5 0

At 10 minutes past 7’o clock the minute hand will be on the number 2.

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A ladder is placed 30 inches from a wall. It touches the wall at a height of 50 inches from the ground.

tankabanditka [31]

To find the length of the ladder, you need to do Pythagorean theorem.
50^2 + 30^2 = x^2
2500 + 900 = x^2
x^2 =3400
x= square root of 3400
58.31 OR 10 root34

To find the angle:
tan theta = opposite/adjacent
50/30 = 5/3
theta = tan inverse of 5/3
= 59.04

8 0

4 years ago

Suppose that a computer software company has 30

LekaFEV [45]

Answer: $\dfrac{30!}{6!(24)!}$

Step-by-step explanation:

Given : The total number of programmers in the company = 30

The company wants to select a group of 6 programmers to work on a particular project.

Since the order of selecting them does not matters , therefore we use combinations.

The number of combinations of r things taken from n things is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

here, n= 30 and r= 6

So the number of different ways to form they could select a group of 6 would be $^{30}C_{6}=\dfrac{30!}{6!(30-6)!}$

$=\dfrac{30!}{6!(24)!}\\\\=\dfrac{30\times29\times28\times27\times26\times25\times24!}{(720)24!}=593775$

i.e. Total ways =593775

In terms of factorials , the number of total ways to form they could select a group of 6 is $\dfrac{30!}{6!(24)!}$ .

7 0

3 years ago

True or false <br> all prime numbers are odd <br> i said false is that correct

Mumz [18]

False

2 is the only even prime number, the rest of the prime numbers are odd.

8 0

4 years ago

From Euler’s relation eiθ = cosθ + isinθ,

adelina 88 [10]

Answer:

(a) The graphic representation is in the attached figure.

(b) $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ .

(c) $\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ .

Step-by-step explanation:

(a) Given a complex number $e^{i\theta}$ we know, from Euler's formula that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ . So, it is not difficult to notice that

$|e^{i\theta}|^2 = \cos^2(\theta)+\sin^2(\theta) =1$

so it is on the unit circumference. Also, notice that the Cartesian representation of the complex number is $(\cos(\theta), \sin(\theta))$ .

Now,

$e^{-i\theta} = \cos(\theta)+i\sin(-\theta) = \cos(\theta)-i\sin(\theta)$ .

Notice that $e^{-i\theta}$ has the same modulus that $e^{i\theta}$ , so it is on the unit circumference. Beside, its Cartesian representation is $(\cos(\theta), -\sin(\theta))$ .

So, the points $(\cos(\theta), \sin(\theta))$ and $(\cos(\theta), -\sin(\theta))$ are symmetric with respect to the X-axis. All this can be checked in the attached figure.