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3 years ago

Please show your work

Mathematics

2 answers:

Tcecarenko [31]3 years ago

8 0

Answer:

(Top right) 4

(middle) 9

(bottom left) 3/1 (milk to eggs)

Step-by-step explanation:

Harrizon [31]3 years ago

6 0

I'll just do the workings,

First Vertical Row

6/2 (as you need to put the milk over the eggs) = 3/1 (simplify the fraction, both can be divided by 2)

= 3/1

Second Vertical Row

= x/3 = 3/1 (x is basically milk for the second vertical row)

= x = 3/1 x 3

= x = 9

Milk = 9

Third Vertical Row

x is the egg,

= 12 / x = 3/1 (3/1 is the same as 3)

= x= 12 / 3

= x = 4

Egg = 4

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Find the inner product for (7, 2) * (0, -2) and state whether the vectors are perpendicular.

Andrews [41]

Answer: A

Step-by-step explanation:

To find the inner product of two vectors (a,b) and (c,d) you would use the equation (a * c) + (b * d)

So for (7,2) and (0,-2) the inner product would be

(7 * 0) + (2 * -2)

= 4

The vectors are only perpendicular when the inner product is equal to 0. Since it is equal to -4 in this case, the vectors are not perpendicular.

A -4; no

3 0

3 years ago

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4 years ago

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X^2+8x= x 2 +8x= \,\,-x −x

hichkok12 [17]

Answer:

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Step-by-step explanation:

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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

Solve the inequality 3x-1>-6x

Ghella [55]

Answer:

x > 1/9

Step-by-step explanation:

Given that:

3x-1>-6x

Adding 6x + 1 on both sides we get

3x-1 + 6x + 1>-6x + 6x +1

3x + 6x > 1

Adding both variables

9x > 1

Dividing both sides b 9 we get

9x/9 > 1/9

x > 1/9

I hope it will help you!

3 0

4 years ago