Ask question

Ask question

All categories

2 years ago

15

Write an equation where the left side is your power of 10 times n and the right side is the result of multiplying 0.1515... by t

hat power.

1 answer:

Neko [114]2 years ago

5 0

Answer:

$n10^x = 0.1515x$

Step-by-step explanation:

Required

Translate the statements to algebraic expression

Represent the power with x;

So, the left hand side is

<em>Power of 10: </em> $10^x$ <em />

<em>Times n: </em> $10^x * n$ <em />

<em />

The right hand side is:

<em>0.1515 * power of 10: </em> $0.1515 * x$ <em />

Equate the right hand side to the left:

$10^x * n = 0.1515 * x$

$n10^x = 0.1515x$

Hence;

<em>The required expression is </em> $n10^x = 0.1515x$ <em />

You might be interested in

A sewer line drops 1 foot every 40 feet. How many feet will it drop over a distance of 100 feet

kodGreya [7K]

Answer:

2.5

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

9. A ladder of length 23 feet leans against the side of a building. The angle of elevation of the ladder is 76. Find the distanc

irakobra [83]

<h3><u>Answer:</u></h3>

B) 22.32 feet

$\quad\rule{300pt}{1pt}$

<h3><u>Solution</u><u>:</u></h3>

we are given that , a ladder is placed against a side of building , which forms a right angled triangle . We wre given one side of a right angled triangle ( hypotenuse ) as 23 feet and the angle of elevation as 76 ° . We can find the Perpendicular distance from the top of the ladder go to the ground by using the trigonometric identity:

$\qquad\quad\bull ~{\boxed{\bf{ Sin\theta =\dfrac{Perpendicular}{Hypotenuse}}} }~\bull$

Here,

hypotenuse = 23 feet
$\theta$ = 76°
Value of Sin $\theta$ = 0.97
Perpendicular = ?

$\quad\dashrightarrow\quad \sf { sin\theta =\dfrac{P}{H}}$

$\quad\dashrightarrow\quad \sf { sin76° =\dfrac{P}{23}}$

$\quad\dashrightarrow\quad \sf { 0.97=\dfrac{P}{23}}$

$\quad\dashrightarrow\quad \sf {P = 0.97 \times 23 }$

$\quad\dashrightarrow\quad \sf { P = 22.32}$

‎ㅤ‎ㅤ‎ㅤ~<u>H</u><u>e</u><u>n</u><u>c</u><u>e</u><u>,</u><u> </u><u>the </u><u>distance </u><u>from </u><u>the </u><u>top </u><u>of </u><u>the </u><u>ladder </u><u>to </u><u>the </u><u>ground </u><u>is </u><u>2</u><u>2</u><u>.</u><u>3</u><u>2</u><u> </u><u>feet </u><u>!</u>

$\rule{300pt}{2pt}$

4 0

2 years ago

Number one solve sin l and tan n cos l and sin n

const2013 [10]

Sin L= 3/5

Tan N= 3/5

Cos L=4/5

Sin N=4/5

5 0

3 years ago

Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

3 years ago

How do I do this?<br> Please help

riadik2000 [5.3K]

<u>Answer </u><u>:</u>

In the given quadrilateral ABCD ,

Angle BCA = 18°
Angle ACD = 62°

Angle BCA = Angle CAD ( alternate interior angle )

Now in triangle CAD ,

We have two angles so by using angle sum property we can find the required third one ,

Angle CAD + Angle ACD + Angle ADC = 180°
18 + 62 + Angle ADC = 180
80 + Angle ADC = 180
Angle ADC = 180 - 80
Angle ADC = 100

6 0

2 years ago

Read 2 more answers