WHAT IS 17 TIMES 38 IN MATH

Mathematics

2 answers:

Stels [109]3 years ago

6 0

17 × 38 =646 :)) this is the answer

Black_prince [1.1K]3 years ago

5 0

♥ 17*38=<span>646</span><span>
To solve this, as it says, you will need to multiply.
Now if you didn't know
the word "times" means to multiply, and multiplying makes this symbol: x or otherwise known as *
♥Hope this helps!♥</span>

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What is when you divide 33-55 divide by 11 will egual

SVETLANKA909090 [29]

So if you means 33-55=-22
-22/11=x

so when dividing by a negative, just divide as usual, but just make the answer negative so

-22/11=22/11 times -1/1=2 times -1=-2

the answer is -2

8 0

3 years ago

Let sin(2x) = cos(x), where 0° ≤ x < 180°. What are the possible values for x?

lions [1.4K]

Answer:

The possible values of x are 90°, 30° and 150°.

Step-by-step explanation:

Given that,

sin(2x) = cos(x) where 0° ≤ x < 180°

We know that, sin(2x) = 2 sinx cosx

2 sinx cosx = cosx

Subtract cosx on both sides

2 sinx cosx - cosx = 0

cosx (2sinx-1)=0

It means, cosx = 0 and (2sin x -1 ) = 0

cos x = cos0 and sinx = 1/2

x = 90° and x = 30°, 150°

Hence, the possible values of x are 90°, 30° and 150°.

6 0

2 years ago

Read 2 more answers

How many elements does A contains if it has 64 subsets

Amanda [17]

<h3>Answer: 6</h3>

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Explanation:

Rule: If a set has n elements in it, then it will have 2^n subsets.

For example, there are n = 3 elements in the set {a,b,c}. This means there are 2^n = 2^3 = 8 subsets. The eight subsets are listed below.

{a,b,c} .... any set is a subset of itself
{a,b}
{a,c}
{b,c}
{a}
{b}
{c}
{ } ..... the empty set

Subsets 2 through 4 are subsets with exactly 2 elements. Subsets 5 through 7 are singletons (aka sets with 1 element). The last subset is the empty set which is a subset of any set. You could use the special symbol $\varnothing$ to indicate the empty set.

For more information, check out concepts relating to the power set.

-------------------

The problem is asking what value of n will make 2^n = 64 true.

You could guess-and-check your way to see that 2^n = 64 has the solution n = 6.

Another approach is to follow these steps.

$2^n = 64\\\\2^n = 2^6\\\\n = 6$

Which is fairly trivial.

Or you can use logarithms to solve for the exponent.

$2^n = 64\\\\\text{Log}\left(2^n\right)=\text{Log}\left(64\right)\\\\n*\text{Log}\left(2\right)=\text{Log}\left(64\right)\\\\n=\frac{\text{Log}\left(64\right)}{\text{Log}\left(2\right)}\\\\n\approx\frac{1.80617997398389} {0.30102999566399} \ \text{ ... using base 10 logs}\\\\n\approx5.99999999999983\\\\$