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frozen [14]
2 years ago
12

What is the total surface area of the rectangular pyramid whose net is shown?

Mathematics
1 answer:
Lunna [17]2 years ago
7 0

Answer:

95 in.²

Step-by-step explanation:

Total surface area of the rectangular pyramid = 4(area of triangle) + area of rectangle

Area of triangle = ½*base*height

base = 5 in.

height = 7 in.

Area of triangle = ½*5*7 = 17.5 in.²

Area of rectangle = Length*width = 5*5 = 25 in.²

Plug in the values

Total surface area of the rectangular pyramid = 4(17.5) + 25

= 95 in.²

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Pls helppp!!! -------
Gelneren [198K]

Answer:

∠ A ≈ 44.42°

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos A = \frac{adjacent}{hypotenuse}  = \frac{AC}{AB} = \frac{5}{7} , then

∠ A = cos^{-1} (\frac{5}{7} ) ≈ 44.42° ( to the nearest hundredth )

5 0
3 years ago
√ 19 + 6^2 ÷ 7.2^2 I need help ASAP
kumpel [21]

Answer:

0.143059384

Step-by-step explanation:

=> \frac{\sqrt{19+6^2} }{7.2^2}

=> \frac{\sqrt{19+36} }{51.84}

=> \frac{\sqrt{55} }{51.84}

=> \frac{7.416}{51.84}

=> 0.143059384

5 0
3 years ago
Read 2 more answers
suppose that the lifetime of a transistor is a gamma random variable x with mean of 24 weeks and standard deviation of 12 weeks.
emmainna [20.7K]

The probability that the transistor will last between 12 and 24 weeks is 0.424

X= lifetime of the transistor in weeks E(X)= 24 weeks

O,= 12 weeks

The anticipated value, variance, and distribution of the random variable X were all provided to us. Finding the parameters alpha and beta is necessary before we can discover the solutions to the difficulties.

X~gamma(\alpha ,\beta)

E(X)= \alpha \beta                 \beta= 12^{2}/24=6 weeks

V(x)= \alpha \beta ^{2}                \alpha=24/6= 4

Now we can find the solutions:

The excel formula used to create Figure one is as follows:

=gammadist(X, \alpha, \beta, False)

P(12\leq X\leq 24)

P(12/6\leq G\leq 24/6)

P(2\leq G\leq 4)

P= 0.424

Therefore, probability that the transistor will last between 12 and 24 weeks is 0.424

To learn more about probability click here:

brainly.com/question/11234923

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4 0
1 year ago
Read 2 more answers
There are approximately 113 million TVs in the United States. Each uses, on average, 76 W of power and is turned on for 6.3 hour
ioda

Answer:

76 W *\frac{1kW}{1000 W}= 0.076 kW

0.076 kW * 6.3 \frac{hours}{day}= 0.4788 \frac{kWh}{day}

113x10^6 people * \frac{0.4788 kWh}{people}= 54104400 kWh

And finally we can convert this into money like this:

54104400 kWh *\frac{0.089 dollars}{1 kWh}= 4815291.6

So they spent every day approximately in total $ 4815291.6 dollars

Step-by-step explanation:

For this case we can convert the average power into Kw like this:

76 W *\frac{1kW}{1000 W}= 0.076 kW

Now on average we know that one person use on average this power over 6.3 hours per day so then we have the following consumption rate:

0.076 kW * 6.3 \frac{hours}{day}= 0.4788 \frac{kWh}{day}

And now we know the total population so then the total consumption per day would be:

113x10^6 people * \frac{0.4788 kWh}{people}= 54104400 kWh

And finally we can convert this into money like this:

54104400 kWh *\frac{0.089 dollars}{1 kWh}= 4815291.6

So they spent every day approximately in total $ 4815291.6 dollars

6 0
3 years ago
Read 2 more answers
Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.
igor_vitrenko [27]

Answer:

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7

So, the solution of the equation is

y = Ae^{m_{1} t} + Be^{m_{2} t}

where m₁ = 3 and m₂ = -7.

So,

y = Ae^{3t} + Be^{-7t}

Also,

y' = 3Ae^{3t} - 7e^{-7t}

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1      (1)

y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}

Substituting A into (1) above, we have

\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1      \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}

Substituting B into A, we have

A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}

Substituting A and B into y, we have

y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

So the solution to the differential equation is

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

6 0
3 years ago
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