All categories

3 years ago

Find T. Pls help!! If you know how to do this and can help me please let me know

Mathematics

1 answer:

Elanso [62]3 years ago

6 0

Answer: t =4.6

Step-by-step explanation:

using sine rule

$\frac{u}{Sin U} = \frac{t}{sin T}$

$\frac{15}{sin 122} = \frac{t}{ sin 15}$

$\frac{15}{0.8480} =\frac{t}{0.2588}$

17.7= $\frac{t}{0.2588}$

cross multiply

17.7 x 0.2588 = t

t=4.58

≈ 4.6

You might be interested in

Tan weighs 146 pounds. Minh weighs 15 pounds more than Tan. How much does Minh weigh?

dimaraw [331]

Given parameters:

Weight of Tan = 146pounds

Unknown:

Weight of Minh = ?

Following this problem step wisely, we can derive a solvable algebraic equation;

Let the weight of Minh = M

In the second sentence, we see that Minh weighs 15pounds more than Tan;

So;

M = 15 + Weight of Tan

Since we already know the weight of Tan,

Weight of Minh = 15 + 146 = 161pounds

We clearly see that Minh's weight will be 161 pounds.

3 0

3 years ago

Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot

Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is:

$\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|$

$\Delta t_{1/2} \approx 65.788\,days$

The approximate difference in the half-lives of the isotopes is 66 days.

4 0

3 years ago

Read 2 more answers

What is the answer to all of these?

4vir4ik [10]

$810000\\
0.000621\\
3040000000\\
0.0000043\\
0.00172\\
58100000\\
0.0965\\
6083$

3 0

3 years ago

HELP PLEASE,<br><br> Part a of the question was posted earlier on.

Elanso [62]

$\bf \lim\limits_{x\to 6}~\cfrac{x^2+x+42}{x-6}\implies \lim\limits_{x\to 6}~\cfrac{(x+7)(x-6)}{(x-6)}\implies \lim\limits_{x\to 6}~x+7
\\\\\\
\lim\limits_{x\to 6}~(6)+7=13
\\\\\\
\textit{since the original equation holds for all, }x\ne 6,\textit{ then it approaches}\\
\textit{the same limit as }x+7$

3 0

2 years ago

For each question, each pair of triangles are similar but not drawn to scale. Calculate any lettered lengths.

Karolina [17]

Answer:

a. 20

b. 5

Step-by-step explanation:

The scale factor is 5 because 6×5=30

a. 4×5=20

b. 25÷5=5

8 0

2 years ago