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Elza [17]
3 years ago
9

I need help for my homework!!

Mathematics
1 answer:
Talja [164]3 years ago
6 0

Answer:

The answer is 4.8cm³

Step-by-step explanation:

First, you have to find the length of BC using trigonometric identities :

sinθ = oppo/hypo

cosθ = adj/hypo

tanθ = oppo/adj

Given that ∠BAC = 45° so you can use tanθ to find length of AB :

tan45° = BC/3.6

BC = 3.6×(tan45°)

= 3.6cm

Now, find the volume of pyramid using V = (1/3)×base area×h :

h = BC = 3.6cm

base area = 2×2

= 4cm²

V = (1/3)×4×3.6

= 4.8cm³

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Hello, can somebody please help me with this word problem? Thanks.
erastovalidia [21]
We are given that this is an ideal gas, and that the volume and presumably the number of moles of gas are constant. We can use Gay-Lussac's Law, which describes volume and pressure. We have that pressure is directly proportional to volume. For a change in a gas, we can write the equation as 
\dfrac{P_i}{T_i} = \dfrac{P_f}{T_f},
where i denotes initial and f denotes final.

We have that T_i=600 \ K, T_f=800 \ K, and P_f=30 \ atm. We need to find P_i. To do so, let's first rearrange Gay-Lussac's equation to solve for P_i.

\dfrac{P_i}{T_i} = \dfrac{P_f}{T_f} \\\\P_iT_f=P_fT_i\\\\P_i= \dfrac{P_fT_i}{T_f}

Now, we plug in our values to get P_i.

P_i= \dfrac{P_fT_i}{T_f}=\dfrac{30\ atm \ 600 \ K}{800 \ K} = 22.5 \ atm.

This seems like a reasonable value, because as temperature goes up, pressure goes up, and an increase in temperature corresponds to an increase in pressure.

Technically, you were given values with only one significant figure, so you can only report the value as 20 \ atm, but this depends on how your instructor usually does these problems!

4 0
3 years ago
CHECK MY ANSWER ASAP
weqwewe [10]

Question 1: Equiangular

Question 2: Area of the parallelogram = 48 square centimeters

Question 3: Perimeter of the rectangle = 42 ft

Question 4: Area of the trapezoid = 154 square inches

Solution:

Question 1:

The given polygon is a rectangle.

The angles of each side of the polygon is 90°.

This means all angles are equal.

Hence Option A equiangular is the correct answer.

Question 2:

Area of the parallelogram = Base × Height]

                                           = 6 cm × 8 cm

Area of the parallelogram = 48 square centimeters

Question 3:

Perimeter of the rectangle = 2(length + width)

                                            = 2( 15 ft + 6 ft)

                                            = 2(21 ft)

Perimeter of the rectangle = 42 ft

Question 4:

Area of the trapezoid = \frac{1}{2}\times \text{sum of the parallel sides}\times \text{height}

                                    $=\frac{1}{2}\times (14+ 8)\times 14

                                    $=\frac{1}{2}\times 22\times 14

Area of the trapezoid = 154 square inches

8 0
4 years ago
What is the area of the trapezoid?
Nana76 [90]

Step-by-step explanation:

A=a+b/2 *h

= 5+11/2*4

32

5 0
3 years ago
Can anyone help me with this question? it is: In 1924 the temperature in Fairfield, Montana, dropped from 63 degrees to -21 degr
Helen [10]
63+12= your difference between the numbers
63+12=84
84/12 = 7.

Your answer would be 7 degrees per hour.
7 0
3 years ago
Figure JKLM is a parallelogram. The measures of line segments MT and TK are shown. What is the value of y?
Katena32 [7]
Because JKLM is a parallelogram, MT = TK.

MT: 8y + 18
TK : 12y - 10

MT = TK
8y + 18 = 12y - 10
8y - 12y = -10 -18
-4y = -28
y = -28/-4
y = 7

MT: 8y + 18 → 8(7) + 18 = 56 + 18 = 74
<span>TK : 12y - 10 </span>→ 12(7) -10 = 84 - 10 = 74

The value of y is 7.
6 0
3 years ago
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