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3 years ago

Help PLZ WITH 24,25 estimate each

Mathematics

1 answer:

arlik [135]3 years ago

5 0

24) $6.85*3 meals=$20.55. No, $20.55 does not equal to $25.42.
25) $92.70/18 working hours=$5.15

No offense, but my 8 year old sister can do this, how are you in high school?

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If TU = 8, UV = 3x, and TV = x + 10, what is TV?

IgorLugansk [536]

The value of TV is 11.

<h3>What is equation?</h3>

An equation is a formula that expresses the equality of two expressions, by connecting them with the equals sign =.

Given:

TU= 8, UV = 3x and TV = x+10

TV= TU + UV

x +10= 8+ 3x

-2x = -2

x= 1

Hence, TV = 1+10= 11.

Learn more about equation here:

brainly.com/question/10413253

#SPJ1

6 0

2 years ago

Please help i really need help

lukranit [14]

9514 1404 393

Answer:

17/99

Step-by-step explanation:

Replace the digits 23 in your example with the digits 17 and you have your answer:

$0.\overline{17}=\dfrac{17}{99}$

_____

In general, a 2-digit repeat will have 99 as its denominator. If the digits are a multiple of 3 or 11, then the fraction can be reduced. 17 is prime, so the fraction cannot be reduced.

3 0

3 years ago

Can anyone help me with this pls? i have no idea what im doing

Musya8 [376]

$h(11)=-20+11\cdot11=-20+121=-101$

8 0

3 years ago

Help me plzz show work plz

iren2701 [21]

Answer:

Step-by-step explanation:

Use the formula for radius and circle circumference.

$d=2r \\ \\ C=2\pi r$

Since we know the diameter, if we divide that by 2, we can get the radius.

$29.5/2=14.75$

Now we can plug that value into the circumference formula to get the circumference.

$C=2(14.75)\pi \\ \\ C=29.5\pi \\ \\ C=93$

*Since we're rounding to the nearest inch.

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

3 years ago