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3 years ago

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Brainlest Subtract the linear expressions. (4/5x - 5) - (1/3x+4) What is the difference? Enter your answer in the box. Enter fra

ctions as simplified fractions.

2 answers:

jekas [21]3 years ago

8 0

Answer:

Step-by-step explanation:

$\left( \dfrac{4}{5}x-5 \right) - \left(\dfrac{1}{3}x+4 \right) = \dfrac{4}{5}x-5-\dfrac{1}{3}x-4\\\\LCM \ of \ 3 \ and \ 5 \ = \ 15\\\\=\dfrac{4*3}{5*3}x - \dfrac{1*5}{3*5}x -9\\\\\\=\dfrac{12}{15}x-\dfrac{5}{12}x-9\\\\=\dfrac{7}{15}x-9$

belka [17]3 years ago

5 0

Answer:

7/15x - 9

Step-by-step explanation:

(4/5x - 5) - (1/3x + 4)

remove the parentheses

4/5x - 5 - 1/3x - 4

simplify each variable

4/5x - 9 - 1/3x

7/15x - 9

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Add.<br> (-3x + 7) + (-6x + 9)

OLEGan [10]

Answer:

<h3>"-9x+16"</h3>

Step-by-step explanation:

(-3x + 7) + (-6x + 9)

<em>SIMPLIFY</em><em>:</em>

<em>=</em><em>-</em><em>3</em><em>x</em><em>+</em><em>7</em><em>+</em><em>-</em><em>6</em><em>x</em><em>+</em><em>9</em><em> </em>(add)

<em>ANSWER</em><em>:</em>

<em>=</em><u><em>-</em><em>9</em><em>x</em><em>+</em><em>1</em><em>6</em></u>

3 0

3 years ago

Read 2 more answers

Z varies jointly with x, and y, and z=7 when x= 2, y= 2<br> pleasee I need an answer now!!

Elina [12.6K]

Answer:

(7/4)(2)(2)=7

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Translate (5, 1) to the right 3 units.

Sergio [31]

Answer:

Step-by-step explanation:

Answer

When you translate either left or right, the x coordinate is the one that you change. To go right when you are dealing with a point, you must add the amount you are asked to go right. So when you go right 3 units, add 3 to the 5.

(5 + 3,1) = (8,1)

when going across the y axis, you are still only changing the x coordinate.

All you need do is put a minus sign in front of the x coordinate. So your final answer is (-8,1)

5 0

2 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago

How do the graphs of y=1/x and y=5/x+6<br><br> compare?

oee [108]

Answer: The graph is attached.

Step-by-step explanation: The given functions whose graphs are to be compared are as follows:

$y=\dfrac{1}{x},~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)\\\\\\y=\dfrac{5}{x}+6.~~~~~~~~~~~~~~~~~~~~~~~~~(B)$

In the attached figure, the graphs of both (A) and (B) are shown. We can easily see see from there, the shapes of both the graphs are same.

But, at x = 0, y = ∞ and at x = ∞, y = 0 in graph (A).

At x = 0, y = ∞ and at x = ∞, y = 6 in graph (B).

Thus, the comparison can be seen in the figure very clearly.

8 0

3 years ago