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3 years ago

1/4(-8x - 12) + -4 = 3/4(-8x - 12)

Mathematics

1 answer:

nekit [7.7K]3 years ago

8 0

Here's our equation:

$\frac{1}{4} (-8x-12)-4=\frac{3}{4} (-8x-12)$

Factor out the left side

(-2x - 3) - 4 = 3/4 (-8x - 12)

Simplify

-2x - 7 = 3/4 (-8x - 12)

Factor out the right side

-2x - 7 = -6x - 9

Add 7

-2x = 6x - 2

Add 6x to isolate the variable

4x = -2

Divide by 4

x = - 1/2

Here is how I simplified both sides initially. I'll only be showing the right side because this is a LOT of typing.

Begin with 3/4 (-8x - 12)

Use rule $a*\frac{b}{c} = \frac{a * b}{c}$ to multiply

3(-8x-12)/4

Factor 3(-8x-12)

(-8x - 12) = (-4 * 2x - 4 * 3)

Simplify (-4 * 2x - 4 * 3)

-3 * 4 (2x+3)

- 12 (2x + 3)

Now that we've factored 3/4 (-8x - 12), we're left with- 12 (2x + 3)/4.

12/4 =3, giving us -3(2x + 3)

Distribute and simplify, then you're done.

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denpristay [2]

Answer:

2.5% and 2.5 · 10^-3, 0.25, 2/5, √5

Step-by-step explanation:

0.25, 2/5, 2.5 · 10^-3, 2.5%, √5

Now let's list them all in the same form, why not decimals.

0.25 = 0.25

2/5 = 4/10 = 40/100 = 0.4

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3 years ago

To complete this component you will need to insert parentheses in Expression 2 to change the order of operations so the expressi

Alika [10]

Given:

The expression is $5+4\times 2+6-2\times 2-1$ .

To find:

The position of parentheses inserted in the expression to get the value 19

Step-by-step explanation:

We have,

$5+4\times 2+6-2\times 2-1$

Now,

$(5+4)\times 2+6-2\times 2-1=9\times 2+6-4-1$

$(5+4)\times 2+6-2\times 2-1=18+6-5$

$(5+4)\times 2+6-2\times 2-1=18+1$

$(5+4)\times 2+6-2\times 2-1=19$

There are more ways to apply the parenthesis, but we do not get 19.

$5+4*\left(2+6\right)-2*2-1=32\neq 19$

$5+4*2+\left(6-2\right)*2-1=20\neq 19$

$5+4*2+6-2*\left(2-1\right)=17\neq 19$

And many more possibilities.

Therefore, the expression after inserting the parentheses is $(5+4)\times 2+6-2\times 2-1$ .

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3 years ago

A consumer bought a CD for $18.25 and paid with a $20 bill. How much change was there?

Lerok [7]

$1.75 would be the answer

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3 years ago

Read 2 more answers

Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

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3 years ago

The number of fives is 3 more than six times the number of tens. Let t represent the number of tens. Write an expression for the

bixtya [17]

Answer:

The required expression is f = 3+6t

Step-by-step explanation:

Let t represent number of tens

Let f represent number of fives

So, the expression for number of fives can be formed by using:

The number of fives is 3 more than six times the number of tens.

f = 3+6t

So, the required expression is f = 3+6t

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3 years ago