Ask question

Ask question

All categories

3 years ago

10

|n/2 − 1/3| −1= 1/6

1 answer:

weqwewe [10]3 years ago

5 0

1. n/2-1/3-1=1/6
n/2=1/6+1/3+1
n/2=1/6+2/6+6/6
n/2=9/6=3/2
n=3/2*2=3

check: abs(3/2-1/3)-1=1/6, abs(9/6-2/6)-1= 1/6, 7/6-1=1/6, 1/6 =1/6
3 is a root

2. -(n/2-1/3)-1=1/6
-n/2+1/3-1=1/6
-n/2-2/3=1/6
-n/2=1/6+2/3
-n/2=1/6+4/6
-n/2=5/6
n=-2*5/6=-5/3=-1 2/3
check : abs((-5/3)/2-1/3) -1=1/6
abs(-5/6-1/3) -1 =1/6, abs(-5/6-2/6) -1=1/6, abs (-7/6)-1=1/6, 7/6-1=1/6, 1/6=1/6
so -5/3 or -1 2/3 is also root

so two roots 3 and -5/3(that can be written as -1 2/3)

You might be interested in

Solve for x x3=8 = x=2 ,x=4,x=24,x=52

Hatshy [7]

$x^3=8\to x=\sqrt[3]8\\\\x=2\ because\ 2^3=8$

3 0

3 years ago

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

3 years ago

Suppose Sally has $28.00 in her bank account and starts saving $18.25 every week. Her friend Suzy has $161.00 in her bank accoun

Elza [17]

a.

Sally has $28.00 in her bank account and starts saving $18.25 every week.

Let x represent the number of weeks from when he have $28.00 in her account.

and y represent the amount in her account at week x;

We can write the function of the case as;

$y=mx+b$

where m is the amount she saves every week, and b the initial amount in her account.

$\begin{gathered} m=\text{ \$18.25} \\ b=\text{ \$28.00} \end{gathered}$

So, the function will be;

$y=18.25x+28$

b.

Her friend Suzy has $161.00 in her bank account and is withdrawing $15 every week.

Using the same rule as in a above;

$\begin{gathered} m=-15\text{ (withdrawing is negative)} \\ b=161 \end{gathered}$

The function will be;

$y=-15x+161$

c.

The graph of the two functions is shown below;

d.

The point where the two lines intersect is shown in the graph above.

The point is at;

$(4,101)$

Which is at week 4 and balance $101.

e.

The point at which they intersect is the point where they both have the same values.

It is the week when both Sally and Suzy have the same amount in there account.

So, at the fourth week (week 4) Sally and Suzy both have $101 in their account

3 0

1 year ago

Describe the transformations of the given function:<br><br> f(x)=−14|x−2|+5

statuscvo [17]

Answer:

im busy so i cant right now

Step-by-step explanation:

7 0

3 years ago

35.6 = the square root of 15.3^2 + the square root of x^2. Find x.

vivado [14]

Answer:

x = 50.9

Step-by-step explanation:

35.6 = √(15.3²) + √(x²)

35.6 = 15.3 + x

x = 35.6 + 15.3

x = 50.9

5 0

3 years ago