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3 years ago

2(x-3)=1/2(4x-12)please help MEH!!!!!....

Mathematics

1 answer:

Zinaida [17]3 years ago

8 0

Answer:

infinitely many solutions

Step-by-step explanation:

2(x-3)=1/2(4x-12)

2x-6=2x-6

both sides are the same so its infinitely many solutions

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An element with mass 800 grams decays by 8.2% per day. How much of the element is remaining after 15 days,

Serjik [45]

Given:

The initial mass of an element is 800 grams.

Decay rate = 8.2% per day

Number of days = 15

To find:

The remaining element after 15 days.

Solution:

The exponential decay model is

$y=a(1-r)^t$

Where, a is the initial value r is the rate of interest and t is time period.

Putting $a=800,r=0.082,t=15$ in the above formula, we get

$y=800(1-0.082)^{15}$

$y=800(0.918)^{15}$

$y=221.68188$

$y\approx 221.7$

Therefore, the mass of the remaining element is 221.7 grams.

7 0

3 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

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