Ask question

Ask question

All categories

3 years ago

13

I F 49x ²b = (7x +1/2)(7x-1/2)then Find the value b.

1 answer:

Marina86 [1]3 years ago

4 0

Answer:

0

Step-by-step explanation:

Apply Only the Outside and Inside Method of the Foil Method.

$7 \times - \frac{1}{2} = - 3.5x$

$\frac{1}{2} \times 7x = 3.5x$

Add them together

$- 3.5x + 3.5x = 0$

So our b value is 0.

You might be interested in

URGENT 50 POINTS WILL MARK BRAINLIEST HELP Mary plans to put carpeting in her house.

fredd [130]

Answer:

225 ft^2

please give branliest if you do so please

Step-by-step explanation:

(12*10)+(5*5)+(8*10)

120+25+50

225

7 0

3 years ago

Read 2 more answers

Hi i need you to convert 9/15 into a decimal and show work please!

kozerog [31]

Answer:

0.6

Step-by-step explanation:

9 is the numerator and 15 is the denominator

divide 9 ÷ 15

which is 0.6

3 0

2 years ago

Read 2 more answers

John took a 5 ½ mile walk to his friend’s house. He left at 11 a.m. and arrived at his friend’s house at 1 p.m. What was John’s

olganol [36]

Answer:

5.5 miles over 2 hours--> 5.5/2= 2.75 miles per hour was his average walking speed. 5.5 miles over 2.5 hours --> 5.5/2.5= 2.2 miles per hour on the return trip meaning that his average speed was (2.75-2.2= 0.55) 0.55 miles per hour lower on the return trip

Step-by-step explanation:

I hope this helps you :)

<em><u>-KeairaDickson</u></em>

7 0

3 years ago

Does anyone know how to solve these

masha68 [24]

Answer:

e=49

f=131

d=90

Step-by-step explanation:

3 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