What is the value of x in the triangle?

Answer:

t

Step-by-step explanation:

A circle can be circumscribed about the given quadrilateral, the answer is: A. True.

We know that,

To circumscribe a circle about a quadrilateral means to place the quadrilateral inside the circle, such that the four vertices of the quadrilateral touches the circumference of the circle.

Opposite angles of a circumscribed polygon are supplementary.

Therefore, a circle can be circumscribed about the given quadrilateral, the answer is: A. True.

Learn more about circumscribed polygon on:

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Ok send the picture so I can help you.

Answer:

what is the question? I can help

Step-by-step explanation:

Answer:

±6

Step-by-step explanation:

Given the mathematical expression;

√36 < √42 < √49

But, √36 = 6 * 6 = 6²

√49 = 7 * 7 = 7²

Simplifying further, we would substitute the new values respectively;

±[6] < √42 < ±[7]

Therefore, evaluating √36 is equal to ±6.

Answer:

a) The total engine volume is 30.3265 in³

b) the mass of air contained in the engine is 0.00129 lbm

c) The efficiency of the engine is 55.6%

Step-by-step explanation:

Given the data in the question;

The configuration is in line with four-stroke, four-cylinder engine.

Bore ( d ) = 3.17 in

Stroke ( L ) = 3.48 in

compression ratio ( r ) = 10.6 : 1

First, we calculate the swept Volume V₁ = π/4 × d²L

we substitute

V₁ = π/4 × (3.17 )² × (3.48) = 27.4655 in³

Now, Compression Ration; r = Vs + Vc / Vc

r = (V₁ + V₂) / V₂

we substitute

10.6 = (27.4655 + V₂) / V₂

10.6V₂ = 27.4655 + V₂

10.6V₂ - V₂ = 27.4655

9.6V₂ = 27.4655

V₂ = 2.861 in³

So

a) total engine volume

[tex]V_{BDC[/tex] = V₁ + V₂

we substitute

[tex]V_{BDC[/tex] = 27.4655 in³ + 2.861 in³

[tex]V_{BDC[/tex] = 27.4655 in³ + 2.861 in³

[tex]V_{BDC[/tex] = 30.3265 in³

Therefore, The total engine volume is 30.3265 in³

b) The mass of air

given that; T = 540°R, P = 14.7 psia

we know that; PV = mRT   { R is constant }

we solve for m

m = PV / RT

we substitute

m = (14.7 × 30.3265)  / ( 0.3704 × 1728 )540

m = 445.79955 / 345627.648

m = 0.00129 lbm

Therefore, the mass of air contained in the engine is 0.00129 lbm

c) The efficiency

from table { properties of air}

At 540°R; u₁ = 92.04 BTu/lbm, u[tex]_r[/tex]₁ = 144.32

Now process 1 - 2 ( Isentropic )   u[tex]_r[/tex]₁ / u[tex]_r[/tex]₂  = V₁/V₁ = r

so,  u[tex]_r[/tex]₁ / u[tex]_r[/tex]₂ = r

144.32  / u[tex]_r[/tex]₂ = 10.6

u[tex]_r[/tex]₂ = 144.32/10.6

u[tex]_r[/tex]₂ = 13.61

Thus, at u[tex]_r[/tex]₂ = 13.61, u₂ = 235 BTU/lbm, T₂ = 1340°R

Now, we know that;

P₁V₁/T₁ = P₂V₂/T₂  ⇒ (P₁/V₁)r = P₂/T₂

⇒ ( 14.7 / 540 )10.6 = P₂/1340

P₂ = 386.664 psia  

process 2 - 3; constant volume heat addition;

Now, At T₃ = 3600°R, u₃ = 721.44 BTU/lbm, u[tex]_r[/tex]₃ = 0.6449

so, [tex]q_{in[/tex] =  u₃ -  u₂

we substitute

so, [tex]q_{in[/tex] =  721.44  -  235  = 486.44 BTU/lbm

Next, process 3 - 4; Isentropic  

u[tex]_r[/tex]₄/u[tex]_r[/tex]₃ = V₄/V₃ = r

u[tex]_r[/tex]₄ / 0.6449  = 10.6

u[tex]_r[/tex]₄ = 6.8359

At u[tex]_r[/tex]₄ = 6.8359; u₄ = 308 BTU/lbm

∴

Heat Rejection [tex]q_{out[/tex] = u₄ - u₁ = 308 - 92.04 = 215.96 BTU/lbm

so, [tex]W_{net[/tex] = [tex]q_{in[/tex] - [tex]q_{out[/tex]

[tex]W_{net[/tex] = 486.44 - 215.96

[tex]W_{net[/tex] = 270.48 BTU/lbm

so the efficiency; η[tex]_{thermal[/tex] = ([tex]W_{net[/tex] / [tex]q_{in[/tex] ) × 100%

η[tex]_{thermal[/tex] = ( 270.48 / 486.44  ) × 100%

η[tex]_{thermal[/tex] = ( 0.55604  ) × 100%    

η[tex]_{thermal[/tex] = 55.6%

Therefore, The efficiency of the engine is 55.6%

Answer:

Answer:

Option B and D are correct

Step-by-step explanation:

Exponential functions are really useful in the real world situation. They can be used to solve following

a) Population Models

b) Determination of area and perimeters

c) Determine time related things such as half life, time of happening of an event etc.

d) Useful for solving financial problems such as computing investments etc.

Hence, option B and D

Answer:

750 were Farm Mechanics.

Step-by-step explanation:

30% of 2500 is 750.

Hope this helps

Answer:

the area is 72

Step-by-step explanation:

Answer: Midpoint: (2, 1)

Step-by-step explanation:

Midpoint formula: ((x1 + x2)/2 , (y1 + y2)/2)

(-4 + 8)/2 = 2

(-8 + 10)/2 = 1

Answer: (2, 1)

Answer:

[tex]sec 0=\frac{\sqrt{66} }{8}[/tex]

Step-by-step explanation:

[tex]Cot0=-4\sqrt{2}[/tex]

[tex]sec0=\frac{\sqrt{33} }{4\sqrt{2} }[/tex]

[tex]=\frac{\sqrt{33} }{4\sqrt{4} } *\frac{4\sqrt{2} }{4\sqrt{2} }[/tex]

[tex]=\frac{4\sqrt{66} }{32}[/tex]

[tex]=\frac{\sqrt{66} }{8}[/tex]

--------------------

hope it helps...

have a great day!!

Answer:

There are 20 odd numbers between 10 and 50

Step-by-step explanation:

Here are all the odd numbers! Hope this helps!

11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49

Answer:

A. I

Step-by-step explanation:

The other person is wrong and I took the test

Cross multiply:

4 x 6 = 9x

24 = 9x

Divide both sides by 9

X = 24/9

X = 2.7 ( rounded to the nearest tenth)

Answer:

umm to me its 2--%456

Step-by-step explanation:

Answer:

no. who eat lunch special = 12

no. who eat dessert = 6

Step-by-step explanation:

Let x = no. who eat lunch special

y = no. who eat dessert

(1)    8x + 2y = 108             (2)    6x + 3y = 90

 Divide thru by 2                         Divide thru by 3

      4x + y = 54                          2x + y = 30

     -2x - y = -30

       2x = 24

         x = 12                               2(12) + y = 30

                                                    24 + y = 30

                                                            y = 6

Step-by-step explanation:

120=2/5t

120×5/2=t

t=600/2

t=300

hope it helps u

Answer: 27

Step-by-step explanation:

Answer:

x5

Step-by-step explanation:

Answer: C.

Step-by-step explanation:

EDGE 2023

Answer:

The wright answer is A

Step-by-step explanation:

[tex] \sqrt{20} = 4.47 \\ \frac{14}{3} = 4.67 \\ 4.47 < 4.5 < 4.67[/tex]

Answer:

a-As the only factor of interest is type of electric motor, thus instead of using 1-factor randomization, a randomized block design is used where at maximum one motor of each type is tested every day.

b- The test statistic is given as

[tex]F_{I-1,(I-1)(J-1)}=\dfrac{12\left(\Sigma^{5}_{i=1}\bar{X}^{2}_{i.}-IJ\bar{X}^2_{..}\right) }{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}[/tex]

Step-by-step explanation:

a-

This is done such that the days are considered as blocks and the randomization is only occuring within the block. The order of testing is random. However the condition is implemented such that one motor of each type is tested each day.

b-

The F-Value is given as

[tex]F-Value=\dfrac{MSA}{MSAB}[/tex]

Here MSA is given as

[tex]MSA=\dfrac{J\Sigma^I_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{I-1}[/tex]

Here

Similarly MSAB is given as

[tex]MSAB=\dfrac{\Sigma^I_{i=1}\Sigma^J_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{(I-1)(J-1)}[/tex]

Here

By putting these values and simplifying, equation becomes:

[tex]F-Value=\dfrac{MSA}{MSAB}\\\\F-Value=\dfrac{\dfrac{J\Sigma^I_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{I-1}}{\dfrac{\Sigma^I_{i=1}\Sigma^J_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{(I-1)(J-1)}}\\\\F-Value=\dfrac{\dfrac{4\Sigma^5_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{5-1}}{\dfrac{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{(5-1)(4-1)}}\\\\[/tex]

This is further simplified as

[tex]F-Value=\dfrac{\dfrac{4\Sigma^5_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{4}}{\dfrac{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{12}}\\\\F-Value=\dfrac{\Sigma^5_{i=1}(\bar{X}_{i.}-\bar{X}_{..})^2}{\dfrac{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{12}}\\\\[/tex]

The numerator can be written as

[tex]F-Value=\dfrac{\Sigma^{5}_{i=1}\bar{X}^{2}_{i.}-IJ\bar{X}^2_{..}}{\dfrac{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}{12}}\\\\F-Value=\dfrac{12(\Sigma^{5}_{i=1}\bar{X}^{2}_{i.}-IJ\bar{X}^2_{..})}{{\Sigma^5_{i=1}\Sigma^4_{j=1}\left(X_{ij}-\bar{X}_{i.}-\bar{X}_{.j}+\bar{X}_{..}\right)^2}}\\[/tex]

Answer: B. 183 cm^3

V = pi r^2 h/3

V = pi (5^2) (7/3)

V = pi (25) (2.33)

V = 183

Answer:

B.183 [tex]cm^{3}[/tex]

Step-by-step explanation:

Volume of cone is [tex]V=\pi r^{2} \frac{h}{3}[/tex]

where h=7 and r=5 (because r=d/2, where d=10)

V=[tex]5^{2}\frac{7}{3}\pi[/tex]

V=25·[tex]\frac{7}{3} \pi[/tex]

V=175/3[tex]\pi[/tex]

V=183.2595715 or 183 [tex]cm^{3}[/tex]

Answer:

Part a:

Since the calculated z= 8.1649 falls in the critical region  z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.

Part b:

P (Type II Error)=  0.6628

Step-by-step explanation:

The null and alternate hypothesis are

H0: μ1 ≥ μ2   against the claim Ha: μ1 < μ2

This is left tailed test

The critical region for this test at 0.10 ∝ is z < ± 1.28

The test statistic

z= x1-x2/ s/√n

z= 66-61/ 3/√24

z= 5/0.61238

z= 8.16486

Since the calculated z= 8.1649 falls in the critical region  z < ± 1.28 we reject H0 and conclude that the assembly time using the new method is faster.

Part b:

P (Type II Error)  = P ( accept H0/ H0 is false)

z ( ∝)= X-u/s/√n

1.28= X-61/0.61238

X= 61.7838

So I will incorrectly fail to reject the null as long as a draw a sample mean that greater than 61.7838

 The probability of a Type II error is given by

P (z> 61.7838-66/24)

P (z>- 0.42161)

From the tables

P (z>- 0.42161)=  0.6628

Answer:

95°

Step-by-step explanation:

95°

They are vertical angles.

Answer:

95 degree write it fast jakakkaoaoa

Answer:

56 m^2

Explanation: L=2, H=2, W=2, d = 6.63324958 m, TSA = 56 m^2, LSA = 48 m^2, TSA = 4 m^2, BSA = 4 m^2, V = 24 m^3

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C. 4.