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Law Incorporation [45]
3 years ago
14

How much times can 9 go into 77

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
6 0
8 times with 5 left over
zavuch27 [327]3 years ago
4 0
8.55 or 8.6 (rounding the 5 up)
You might be interested in
Pls answer asap 7 - 3 r = r - 4 ( 2 + r)​
Marianna [84]

Answer:

no real answer

Step-by-step explanation:

distribute the -4 and combine all of the like terms on the left side = r-8-4r, then -8-3r

then we have 7-3r=-8-3r

from here, we can already tell that there's no real answer. this is because the two -3r will cancel, leaving no variable.

since 7 doesn't equal -8, there is no answer.

if, for example, the value on both sides of the equal sign were the same after the variable was eliminated, then your answer would be all real numbers

8 0
2 years ago
Please help me :((!!!!
Alika [10]
This seems like a concept you’re going to learn and use so I’ll do a problem per section and explain.

Well basically this is a concept: if it’s a negative exponent it’s just 1/number^exponent.

So for example:
#1. 1/(10)^2 = 10^-2

The standard notation:
#9: You know that 1/(10)^3 you could either put it in a calculator or realize that you take the number 1, and move the decimal to the left three times. 0.001 would be the answer.

Scientific notation:
combines these topics. Let’s take #17. 0.025. Scientific notation means you write it as a number multiplied by 10^? So let’s see how many places you can move the decimal to get a number without the zeros.

0.025 => move it two places to the right. So that’s 2.5. Now 2.5 multiplied by what 10^? Would give you 0.025? It would be 10^-2. So your answer would be 2.5 x 10^-2.

For all of these, use your knowledge of the power of tens and dividing by tens! ^^ you got this. Let me know if you need help after this explanation.
7 0
3 years ago
Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6
san4es73 [151]

Answer:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........

The interval of convergence is:(-\frac{2}{3},\frac{16}{3})

Step-by-step explanation:

Given

f(x)= \frac{9}{3x+ 2}

c = 6

The geometric series centered at c is of the form:

\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.

Where:

a \to first term

r - c \to common ratio

We have to write

f(x)= \frac{9}{3x+ 2}

In the following form:

\frac{a}{1 - r}

So, we have:

f(x)= \frac{9}{3x+ 2}

Rewrite as:

f(x) = \frac{9}{3x - 18 + 18 +2}

f(x) = \frac{9}{3x - 18 + 20}

Factorize

f(x) = \frac{1}{\frac{1}{9}(3x + 2)}

Open bracket

f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}

Rewrite as:

f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}

Collect like terms

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}

Take LCM

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}

f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}

So, we have:

f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}

By comparison with: \frac{a}{1 - r}

a = 1

r = -\frac{1}{3}x + \frac{7}{9}

r = -\frac{1}{3}(x - \frac{7}{3})

At c = 6, we have:

r = -\frac{1}{3}(x - \frac{7}{3}+6-6)

Take LCM

r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}ar^n

Substitute 1 for a

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}1*r^n

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}r^n

Substitute the expression for r

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n

Expand

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]

Further expand:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................

The power series converges when:

\frac{1}{3}|x - \frac{7}{3}| < 1

Multiply both sides by 3

|x - \frac{7}{3}|

Expand the absolute inequality

-3 < x - \frac{7}{3}

Solve for x

\frac{7}{3}  -3 < x

Take LCM

\frac{7-9}{3} < x

-\frac{2}{3} < x

The interval of convergence is:(-\frac{2}{3},\frac{16}{3})

6 0
2 years ago
4x &lt; 8<br> And how to graph it?
Montano1993 [528]

Answer:

x<2

put a circle (unsahded) on number 2 and a line on all the sides to the left of 2 , like through 1, 0 , -1 ....

Step-by-step explanation:

4x<8

divide by 4

x<2

8 0
3 years ago
Which fraction has a repeating decimal as its decimal expansion?<br> 3/25<br> 3/16<br> 3/11<br> 3/8
USPshnik [31]
It should be 3/11 because we know that 11 is a number where if almost any number divide 11 is a repeating.
3 0
2 years ago
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