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Ilia_Sergeevich [38]
4 years ago
12

What number squared is 27?

Mathematics
2 answers:
alexdok [17]4 years ago
6 0
5.20 squared is 27.1 is that close enough?

Lunna [17]4 years ago
4 0
The answer is 5.19615242271 but if you round it to the nearest dp its 5.2
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What is <br><img src="https://tex.z-dn.net/?f=f%284%29%20%3D%20%20%5Csqrt%7Bx%20%7B%7D%5E%7B2%7D%20%7D%20%20-%205" id="TexFormul
dexar [7]
<h3>Solution:</h3>

f(x) =  \sqrt{ {x}^{2} }  - 5 \\  =  > f(4) =  \sqrt{ {(4)}^{2} } - 5 \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \sqrt{16 }   - 5 \\  =  4 - 5 \\  =  - 1 \\ or \\  =  - 4 - 5 \\  =  - 9

<h3>Answer:</h3>

-1 or -9

<h3>Hope it helps</h3>

ray4918 here to help

3 0
3 years ago
Need help with this. Hope someone can help
ratelena [41]

First, convert all of the cm measurements to m measurements (so they are all the same unit measurement)

2000 cm = 20 m         800 cm = 8m    

<u>Total Perimeter </u>(Note that circumference of a semi-circle is 2 π r/2 = π r)

Add up the lengths of all of the outside edges. I am going to start on the top and move counter-clockwise:

40 + π (10) + 8 + 25 + 8 + (40 - 25 - 10) + 8 + 10 + 8 + π(10)

= 40 + 10π + 41 + (5) + 26 + 10π

= 112 + 20π

= 112 + 62.8

= 174.8

Answer: 174.8 m

<u>Total Area</u>

Split the picture into 5 sections (2 semi-circles, top rectangle, bottom left rectangle, and bottom right rectangle).  Find the area for each of those sections and then add their areas together to find the total area.

2 semi-circles is 1 Circle: A = π · r²  ⇒ A = π(20/2)² =  π(10)²  = 100π  ≈  314

top rectangle: A = L x w  ⇒  A = 40 x 20  = 800

bottom left rectangle: A = L x w  ⇒  A = 25 x 8  = 200

bottom right rectangle: A = L x w  ⇒  A = 10 x 8  = 80

Total = 314 + 800 + 200 + 80  = 1394

Answer: 1394 m²


3 0
4 years ago
on a coordinate grid wgat is the horizontal and the vertical line called... I just cant bother to look in my notes
ratelena [41]
Horizontal is called the x-axis and the vertical is called the y-axis
7 0
4 years ago
Read 2 more answers
PLS HELP ITS MY FINAL EXAM What is the area of the base of the cone below? A cone with height 12 feet and volume 90 feet cubed.
hoa [83]

Answer:

A=22.5\ \text{feet}^2

Step-by-step explanation:

We have,

Height of a cone is 12 feet

Volume of the cone is 90 feet

It is required to find the area of the base of the cone.

The base of a cone is circular in shape.

The formula used to find the volume of a cone is given by :

V=\dfrac{1}{3}\pi r^2 h

here, \pi r^2=A (area of base of cone)

V=\dfrac{1}{3}\times Ah\\\\A=\dfrac{3V}{h}\\\\A=\dfrac{3\times 90}{12}\\\\A=22.5\ \text{feet}^2

So, the area of the base of the cone is 22.5\ \text{feet}^2.

3 0
3 years ago
Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground
Scorpion4ik [409]

Answer:

125.4\ \text{m/s}

Step-by-step explanation:

u = Initial velocity of baseball

\theta = Angle of hit = 30^{\circ}

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

y_0 = Height of hit = 2.5 ft

a_y = g = Acceleration due to gravity = 32.2\ \text{ft/s}^2

t = Time taken

Displacement in x direction

x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}

Displacement in y direction

y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}

The minimum initial velocity needed for the ball to clear the fence is 125.4\ \text{m/s}

5 0
3 years ago
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