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Alex73 [517]
2 years ago
6

What effect would doubling all the dimensions of a triangular pyramid have on the volume of the pyramid?

Mathematics
1 answer:
lana66690 [7]2 years ago
3 0

Answer:

The effect of doubling all the dimensions of a triangular pyramid will have on the volume of the pyramid is to increase it by 8 times.

Step-by-step explanation:

i) volume of triangular pyramid  = \frac{1}{3} \times Area \times height

     = \frac{1}{3} \times (Base of triangle \times perpendicular height of triangle) \times height

ii) if we double all the dimensions then the three variables((Base of triangle, perpendicular height of triangle, height of pyramid) will be doubled

  and the volume of the new pyramid will be 8 times that of the original one.

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Factor the following trinomials . <br><br> x2 + 2x - 35 <br><br> x2 - x - 12
dexar [7]
X2-2x-35=0x2-7x+5x-35=0x(x-7)+5(x-7)=0(x-7)(x+5)=0x-7=0      or  x+5=0x=7         or  x=-5 Check: x=7 ,LHS = 49-14-35                          =0 =RHS           x=-5, LHS = 25+10-35                           =0=RHSAns: x=7 or x=-5


<span><span>x2</span>−x−12=0</span>


6 0
3 years ago
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I suck at math and i really need help asap!
AlladinOne [14]

Answer:

2

Step-by-step explanation:

2^{4-3}

=2

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Vincent mows 8 lawns in 2 hours. How many lawns are mowed in 5 hours?
Alex777 [14]
<h3>Answer: 20 lawns</h3>

Work Shown:

8 lawns = 2 hours

4 lawns = 1 hour (divide both sides by 2)

20 lawns = 5 hours (multiply both sides by 5)

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3 years ago
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Not sure what to do here can someone please help
kumpel [21]

Answer:

  x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

__

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

  \sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

  x = -1: √(-1+2) +1 = √(3(-1)+3)   ⇒   1+1 = 0 . . . . not true

  x = 2: √(2+2) +1 = √(3(2) +3)   ⇒   2 +1 = 3 . . . . true . . . x = 2 is the solution

__

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

  u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}

Solutions to this equation are ...

  u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

  x = u² -2 = 2² -2

  x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

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1 year ago
Why isn't 10 a perfect square
nadya68 [22]
Because there is no number that when multiplied by itself that could equal to 10 (ex: 16 is a perfect square because 4 and 4 multiplied together equals 16; 4 and 4 are the same number)
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