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2 years ago

Nine times a number decreased by four

Mathematics

2 answers:

Wewaii [24]2 years ago

7 0

The answer for this question is 9x-4

PolarNik [594]2 years ago

5 0

Answer: 9x - 4

Step-by-step explanation:

We know that a number must be decreased by four. Let's break it up a little.

Nine times <em>some </em>number. We don't know what nine will be multiplied by, so we will just write a placeholder number: x. This number will be multiplied by nine. So it will turn into 9x.

We must decrease (subtract) from this by four. Subtracting 9x by four can be written as 9x - 4.

You might be interested in

What is the answer to (8y -6

scZoUnD [109]

Answer:

Step-by-step explanation:

Incomplete presentation of question. Did you want to factor out the greatest common factor (which is 2), obtaining 2(4y - 3)? Or something else? Also: please note that if you use the left parenthesis ) , you MUST also use the right parenthesis ( .

6 0

3 years ago

Which expression is equivalent to ^

mash [69]

Answer:

Step-by-step explanation:

lets break down 120 to see what it is divisible by and see if any of those numbers are perfect squares.

120/2=60

120/3=40

120/4=30

we can stop there because 4 is a perfect square and 30 can not be reduced any further to produce a perfect square.

do not forget there is only one x so it must stay in side the radical.

your answer is 2sqrt(30x)

5 0

3 years ago

a business want to increase its number of employees by 65% to 20,460. How many employees does the company currently have?

Crazy boy [7]

Answer:

12,400

Step-by-step explanation:

Let the current employees be x

65% increase in number = 0.65x

If the total number of employees after the increase is 20,460, then;

x + 0.65x = 20,460

1.65x = 20,460

x = 20,460/1.65

x = 12,400

Hence the current number of employees is 12,400

5 0

3 years ago

Is there an ordered pair solutions to <br> -3x+3y=4 and <br> -x+y=3 ????

elixir [45]

Y=x+3
-3x+3(x+3)=4
-3x+3x+9=4
9=4
Since this statement is false, there are no ordered pair solutions to this problem.

8 0

3 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