Find the polynomial that represents this figure:

A sample of radium has a weight of 1.5 mg and decays

Oliga [24]

Answer: $(b)$

Step-by-step explanation:

Given

Initial mass of radium is $A_o=1.5\ mg$

It decays by half every 6 years i.e. $T_{\frac{1}{2}}=6$

For radioactive decay, sample at any time is given by

$\Rightarrow A=A_o2^{-\dfrac{t}{T_{\frac{1}{2}}}}$

Insert the values

$\Rightarrow A=1.5\cdot 2^{\dfrac{-t}{6}}\\\\\Rightarrow A=1.5\cdot \left(\dfrac{1}{2}\right)^{\dfrac{-t}{6}}\\\\\Rightarrow A=1.5\cdot (0.5)^{\dfrac{t}{6}}$

Thus, option (b) is correct.

6 0

3 years ago

Match the equation of circle with its center and radius.

svlad2 [7]

Answer:

1- c=(-2,1) , R =9

2-c=(2,-1) , R=3

3-c=(-2,-1) , R=4

4-c=(2,1) , R=16

5-c=(0,0) , R=2

6 0

3 years ago

Whats 9 +10 will give brainlest

shtirl [24]

Answer:

19

Step-by-step explanation:

"You stupid"

"No Im not"

"Whats 9+10"

"21"

"You stupid"

4 0

3 years ago

Read 2 more answers

What is the equation of a line that is perpendicular to −x+3y=9 and passes through the point (−3, 2) ?

vekshin1

The equation of line that is perpendicular to the line $-x+3y =9$ and passes through the point $\left({-3,2}\right)$ is $\boxed{{\mathbf{y=-3x-7}}}$ .

Further explanation:

It is given that the equation of line is $-x+3y =9$ and passes through point $\left({-3,2}\right)$ .

Rewrite the given equation $-x+3y =9$ as follows:

$\begin{aligned}-x+3y&=9\\3y&=9+x\\y&=\frac{9}{3}+\frac{1}{3}x\\y&=\frac{1}{3}x+3\\\end{aligned}$

Now, compare the obtained equation of line $y=\frac{1}{3}x+3$ with the standard equation of line $y=mx+b$ .

$\begin{aligned}m&=\frac{1}{3}\\b&=3\\\end{aligned}$

Therefore, the slope is $\frac{1}{3}$ .

It is given that both lines are perpendicular to each other so the product of slope must be equal to $-1$ .

${m_1}\cdot {m_2}=-1$ …… (1)

Substitute $\frac{1}{3}$ for ${m_1}$ in equation (1) to obtain the value of slope ${m_2}$ .

$\begin{aligned}\frac{1}{3}\cdot{m_2}&=-1\\{m_2}&=-3\\\end{aligned}$

Therefore, the slope is $-3$ .

It is given that the line passes through point $\left({-3,2}\right)$ .

The point-slope form of the equation of a line with slope $m$ passes through point $\left({{x_1},{y_1}}\right)$ is represented as follows:

$y-{y_1}=m\left({x-{x_1}}\right)$ …… (2)

Substitute $-3$ for ${x_1}$ , $2$ for ${y_1}$ and $-3$ for $m$ in equation (2) to obtain the equation of line.

$\begin{aligned}y-2&=-3\left({x-\left({-3}\right)}\right)\\y-2&=-3\left({x+3}\right)\\y&=-3x-9+2\\y&=-3x-7\\\end{aligned}$

Therefore, the equation of line is $y=-3x-7$ .

Thus, the equation of line that is perpendicular to the line $-x+3y=9$ and passes through the point $\left({-3,2}\right)$ is $\boxed{{\mathbf{y=-3x-7}}}$ .

Learn more:

1. Which classification best describes the following system of equations? brainly.com/question/9045597

2. What is the value of $x$ in the equation $x-y=30$ when $y=15$ ? brainly.com/question/3965451

3. What are the values of x? brainly.com/question/2093003

Answer Details:

Grade: Junior High School

Subject: Mathematics

Chapter: Coordinate Geometry

Keywords: Coordinate Geometry, linear equation, system of linear equations in two variables, variables, mathematics, equation of line, line, passes through point.

6 0

4 years ago

Read 2 more answers

Find the area of the shape shown below.

Kobotan [32]

Area=28cm

Explanation:

4 0

4 years ago