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Ira Lisetskai [31]
3 years ago
13

Find the surface area of square pyramid someone help

Mathematics
1 answer:
IgorC [24]3 years ago
4 0

Answer:

65 square units

Step-by-step explanation:

The base of the pyramid is 5 by 5 , surface area is 25 square units.

the side of each one is 4 units "high" and 5 long at the base and to find the are of a triangle you multiple base times height then dived by 2. Therefore each side of the triangle is 10 units square. You have 4 sides so...

40 (total of all the sides ) +25 ( base ) = 65

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Prime factorisation of 500​
Andrews [41]

\boxed{\underline{\bf \: ANSWER}}

2 <u>| 500</u>

2 <u>| 250</u>

5 <u>| 125</u>

5 <u>| 25</u>

5 <u>| 5</u>

1 <u>| 1</u>

Prime factorisation of <u>500 = 2 × 2 × 5 × 5 × 5 × 1</u>

_____

Hope it helps.

ᏒᎯᎨᏁᏰᎾᏯᏕᎯᏝᎿ2222

6 0
2 years ago
Solve the system of equations by graphing: -1/3x + y=-1, y=4+1/3x
Arada [10]

-\frac{1}{3}x + y = -1   ⇒   y = \frac{1}{3}x - 1

To graph this line, plot a point at the y-intercept (0, -1), than plot the next point using the rise over run from the slope (\frac{1}{3}) by counting up 1 and to the right 3 of the y-intercept.  This gives you a second point of (3. 0).  Draw a line through those two coordinates.

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***************************************************************************************

y = 4 +  \frac{1}{3}x    ⇒   y =  \frac{1}{3}x  + 4

Same as above.  Plot the y-intercept (0, 4) and then use rise over run from the slope to plot (3, 5).

Answer: Plot (0, 4) and (3, 5) and draw a line through them.

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You should end up with two PARALLEL lines.  Since the lines never intersect, there are no solutions to this system of equations.

Answer: No Solution

3 0
3 years ago
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Answer:

-2x-5y+8z+4.5=0

Step-by-step explanation:

Let (x,y,z) be the coordinates of the point lying on the needed plane. This point is equidistant from the points (-3, 5, -4) and (-5, 0, 4), so

d_1=\sqrt{(x-(-3))^2+(y-5)^2+(z-(-4))^2}=\sqrt{(x+3)^2+(y-5)^2+(z+4)^2}\\ \\d_2=\sqrt{(x-(-5))^2+(y-0)^2+(z-4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\d_1=d_2\Rightarrow \sqrt{(x+3)^2+(y-5)^2+(z+4)^2}=\sqrt{(x+5)^2+y^2+(z-4)^2}\\ \\(x+3)^2+(y-5)^2+(z+4)^2=(x+5)^2+y^2+(z-4)^2\\ \\x^2+6x+9+y^2-10y+25+z^2+8z+16=x^2+10x+25+y^2+z^2-8z+16\\ \\-4x-10y+16z+9=0\\ \\-2x-5y+8z+4.5=0

5 0
3 years ago
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