Hey please I need help with this question<br><br> 2)

Help! what is the surface area of the prism? 128in^2, 768in^2, 384in^2, or 480in^2

Nimfa-mama [501]

Option C: $384 \ in^2$ is the surface area of the prism

Explanation:

Given figure is a triangular prism.

We need to determine the surface area of the triangular prism.

The formula to find the surface area of the triangular prism is given by

$S A=b h+(s 1+s 2+s 3) H$

where b is the base of the triangle,

h is the height of the triangle,

$s1,s2,s3$ are the three sides of the triangle and

H is the height of the prism.

Let us substitute the values $b=16$ , $h=6$ , $s1=16$ , $s2=10$ , $s3=10$ and $H=8$ in the formula, we get,

$SA=16(6)+(16+10+10)(8)$

Simplifying, we get,

$SA=96+(36)(8)$

Multiplying the term, we get,

$SA=96+288$

Adding the terms, we have,

$SA=384 \ in^2$

Thus, the surface area of the prism is $384 \ in^2$

Hence, Option C is the correct answer.

6 0

3 years ago

The graph below represents the amount of water measured in a swimming pool as a function of time.?

Anastaziya [24]

Answer:

The graph shows that

As time increases, the amount of water in the pool also increases

3 0

2 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

The length of a rectangle is 3 times its width. If the length were increased by 2 inches and the width by 1 inch, the new perime

Dennis_Churaev [7]

Pretty sure it’s the second option because length would be 3w+2
and width would be w+1 and with area you have to multiply so it would just be the (w+1)(3w+2)

5 0

2 years ago

Read 2 more answers

How do you write the equation 0=x+1 in standard form

Elanso [62]

Answer:

x=-1

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Hey please I need help with this question 2)