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3 years ago

What position is the 9 in the number 9,32805

2 answers:

7 0

( I think there's a typo. The comma should be in between 2 and 8 )

The 9 is in the hundred thousands place.

932805
5- ones
0- tens
8-hundreds
2-thousands
3-ten thousands
9-hundred thousands

Yanka [14]3 years ago

5 0

Solutions

To solve the problem you can you place value.

The number 9 is in the hundred thousands place.

= Number (932805)

5- ones

0- tens

8-hundreds

2-thousands

3-ten thousands

<span>9-hundred thousands
</span>

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Please show ALL work!<br><br> Solve for x: log₂x+log₂(x-6)=4

Vlad [161]

Answer: x = 8

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I'm going to use the notation log(2,x) to indicate "log base 2 of x". The first number is the base while the second is the expression inside the log (aka the argument of the log)

log(2,x) + log(2,(x-6)) = 4
log(2,x*(x-6)) = 4
x*(x-6) = 2^4
x*(x-6) = 16
x^2-6x = 16
x^2-6x-16 = 0
(x-8)(x+2) = 0
x-8 = 0 or x+2 = 0
x = 8 or x = -2

Recall that the domain of log(x) is x > 0. So x = -2 is not allowed. The same applies to log(2,x) as well.

Only x = 8 is a proper solution.

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

You can use the change of base rule to check your work
log base 2 of x = log(2,x) = log(x)/log(2)
log(2,(x-6)) = log(x-6)/log(2)

So,
(log(x)/log(2)) + (log(x-6)/log(2)) = 4
(log(8)/log(2)) + (log(8-6)/log(2)) = 4
(log(8)/log(2)) + (log(2)/log(2)) = 4
(log(2^3)/log(2)) + (log(2)/log(2)) = 4
(3*log(2)/log(2)) + (log(2)/log(2)) = 4
3+1 = 4
4 = 4
The answer is confirmed

5 0

3 years ago

What is the degree of y= x2

IRISSAK [1]

Answer:

2nd degree

Step-by-step explanation:

4 0

3 years ago

Prove 2^n > n for all n equal to or greater than 1. I mostly need help with how to solve the problem when it is greater than

noname [10]

If $n$ is an integer, you can use induction. First show the inequality holds for $n=1$ . You have $2^1=2>1$ , which is true.

Now assume this holds in general for $n=k$ , i.e. that $2^k>k$ . We want to prove the statement then must hold for $n=k+1$ .

Because $2^k>k$ , you have

$2^{k+1}=2\times2^k>2k$

and this must be greater than $k+1$ for the statement to be true, so we require

$2k>k+1$

for $k>1$ . Well this is obviously true, because solving the inequality gives $3k>1\implies k>\dfrac13$ . So you're done.

If you $n$ is any real number, you can use derivatives to show that $2^n$ increases monotonically and faster than $n$ .

7 0

3 years ago

The speeds of cars on a given i street we are normally distributed with a mean of 72 miles per hour and a standard deviation of

Ostrovityanka [42]

Answer: 72.78% of the drivers are traveling between 70 and 80 miles per hour based on this distribution.

Step-by-step explanation:

Let X be a random variable that represents the speed of the drivers.

Given: population mean : M = 72 miles ,

Standard deviation: s= 3.2 miles

The probability that the drivers are traveling between 70 and 80 miles per hour based on this distribution:

$P(70\leq X\leq 80)=P(\frac{70-72}{3.2}\leq \frac{X-M}{s}\leq\frac{80-72}{3.2})\\\\= P(-0.625\leq Z\leq 2.5)\ \ \ \ \ [Z=\frac{X-M}{s}]\\\\=P(Z\leq2.5)-P(Z\leq -0.625)\\\\\\ =0.9938-0.2660\ \ \ [\text{Using p-value calculator}]\\\\=0.7278$

Hence, 72.78% of the drivers are traveling between 70 and 80 miles per hour based on this distribution.

3 0

3 years ago

There are 17 kites in a kite-flying contest at the beach. There are 3 times as many box kites as butterfly kites. There are 2 mo

alina1380 [7]

C is your answer. 5x + 2 = 17

First let’s get them all in terms of x:
Box kites = 3x
Dragon kites = x + 2
Butterfly kites = x

Adding all those up:

x + 3x + x + 2 = 17
Combine like terms:
5x + 2 = 17

6 0

3 years ago

Read 2 more answers