All categories

3 years ago

What does 123 times 504 equal?? I am just soooo curious about this... ;)

Mathematics

2 answers:

tatiyna3 years ago

6 0

Answer:

61992

Step-by-step explanation:

timofeeve [1]3 years ago

6 0

Answer:

$\huge\boxed{Answer\hookleftarrow}$

$123 \times 504 \\ = 61992$

You might be interested in

Which ratios are equal to each other? A. 1 : 16 and 2 : 8 B. 2 : 32 and 4 : 8 C. 1 : 64 and 4 : 16 D. 1 : 16 and 4 : 64

Ronch [10]

The answer to the question is d

5 0

3 years ago

Read 2 more answers

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

3 years ago

Please answer correctly

TiliK225 [7]

Answer:

Step-by-step explanation:

Since side BC = side AD (because opposite side of a parallelogram are congruent)

Angle C = angle A(because opposite angles of a parallelogram are congruent)

side DC = side AB(because the opposite sides of a parallelogram are congruent. Therefore,

Triangle ABD is congruent to triangle CDB by the postulate of SAS which means side angle side.

3 0

3 years ago

Pleeeeease help!!!!!!!!

Gnom [1K]

Answer with explanation:

Data set Showing number of times seventh period class attended a summer camp since first grade:

3, 1, 0, 4, 1,⇒⇒Mean =1.8

4, 2, 3, 0, 5,⇒⇒Mean =2.8

0, 1 , 1, 2, 0,⇒⇒Mean =0.8

4, 1, 4, 4, 2,⇒⇒Mean =3

3, 2, 0, 1, 4⇒⇒Mean =2

⇒≡Mean of a Data set is sum of all the variate in data set divided by total number of Variate in the data set.

Range of the sample Means of a data set is difference between Maximum value of variate in the data set and Minimum value of Variate in Data set.

Maximum Value in the sample mean = 3

Minimum Value in the sample mean = 0.8

⇒Range = Maximum value of mean - Minimum value of mean

= 3 - 0.8

= 2.2

⇒≡None of the option given, matches with the value of range Calculated.

8 0

3 years ago

Read 2 more answers

Which relation is a function

Artemon [7]

I think that is the second one

5 0

3 years ago