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Oksanka [162]
3 years ago
15

Help asap geometry 1 honors!!!!

Mathematics
1 answer:
Elis [28]3 years ago
6 0

Answer: 12

Step-by-step explanation:

42

42

42

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1/4x +4 = 1/2 solve with explanation
IRINA_888 [86]

Answer:

<h3>x = -14</h3>

Step-by-step explanation:

Given the equation

1/4x +4 = 1/2

Solving for x

Subtract 4 from both sides

1/4x +4 - 4 = 1/2 - 4

1/4 x  = (1-8)/2

1/4 x = -7/2

Cross multiply

2x = -7*4

2x = -28

x = -28/2

x = -14

Hence the value of x is -14

3 0
3 years ago
PLEASE HELP ASAP, I HAVE 10 MINUTES ;
Ilya [14]

Answer:

Step-by-step explanation:

when 13 bought we find these inequalities

x + 12y < 7  no

2x + 11y < 7.5 no

3x + 10y < 8 no

4x + 9y < 8.5 no

5x + 8y >  9  yes

When 14 bought we find these inequalities

x + 13y < 7.5 no

2x + 12y < 8 no

3x + 11y < 8.5 no

4x + 10y > 9 yes  

7 0
3 years ago
If f (x) = 2 x + 5 and three-halves are inverse functions of each other and f(x)=2x+5, what is f-1(8)?
nalin [4]

Answer:

b

Found the answer, its B. 3/2!

6 0
3 years ago
Calculate the area of the shaded region
velikii [3]

Answer:

48π or 150.72

Step-by-step explanation:

The radius of the big circle is 8.

We can use that to find the area of the big circle.

Area is πr^2

π(8)^2

64π

Now, we can find the area of the small circle.

The diameter of that is 8, meaning the radius is 4.

πr^2

π(4)^2

16π

Now we subtract the area of the small circle from the area of the big circle to get the area of the shaded region.

64π-16π=48π

As a decimal: 48*3.14=150.72

4 0
3 years ago
A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave
erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, L = 120 cm

Its wavelength \lambda_1 = 480 cm

                         \lambda_2 = 160 cm and \lambda_3 = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

  • The fundamental wavelength of a pipe which is open at both ends is 2L.
  • The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ 4L=4(120)=480\ cm

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒  \lambda_2=\frac{4L}{3}                                   ⇒  \lambda_3=\frac{4L}{5}

⇒ \lambda_2=\frac{4(120)}{3}                              ⇒  \lambda_3=\frac{4(120)}{5}

⇒ \lambda_2=\frac{480}{3}                                  ⇒  \lambda_3=\frac{480}{5}

⇒ \lambda_2=160\ cm                           ⇒   \lambda_3=96\ cm  

So the pipe is open at one end and closed at one end .

6 0
3 years ago
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