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Katarina [22]
2 years ago
8

How many bases are there in the triangular pyramid shown below?

Mathematics
2 answers:
sukhopar [10]2 years ago
5 0

Answer:

1

Step-by-step explanation:

muminat2 years ago
5 0

Answer:

1

Step-by-step explanation:

In a pyramid, there is only one base that touches the ground.

However, in a prism, there are two bases, two of the same ends.

The front of the name of the shape also hints:

If it's an octagonal prism, there are two octagon bases.

If it's a hexagonal pyramid, there is one hexagon base.

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She has to buy both binders and notebooks. So, you have to take into account that she has to have both. The closest you can get to $20 while still getting notebooks, is to buy 4 binders. 4 times 4 equals 16. So, she can get 4 binders and 2 notebooks, because then, 2 times 2 equals 4 and 16 plus 4 equals 20.
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Heights in inches of several tomato plants : 16,18,18,20,17,10,18,17 mean absolute deviation :
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18 would have to be the mean because of the average
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3 years ago
Could someone help me ASAP, thanks !!
Rashid [163]

Answer:

I think this is the right answer but last two signs are incorrect

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3 years ago
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How do I solve for x
Alchen [17]
Greetings!

To find the length of any side of a right triangle, you can use the Pythagorean Thereom. It states that the squares of two sides are equal to the square of the hypotenuse:
a^2+b^2=c^2

Input the information from the diagram into the formula: 
(x)^2+(x+7)^2=(13)^2

Expand each term:
(x)^2+(x+7)^2=(13)^2

x^2+((x+7)(x+7))=169

x^2+(x(x+7)+7(x+7))=169

x^2+x^2+7x+7x+49=169

Combine like terms:
2x^2+14x+49=169

Add -169 to both sides:
(2x^2+14x+49)+(-169)=(169)+(-169)

2x^2+14x-120=0

Factor out the Common Term (2):
2(x^2+7x-60)=0

Factor the Complex Trinomial:
2(x^2-5x+12x-60)=0

2(x(x-5)+12(x-5))=0

2(x-5)(x+12)=0

Set Factors to equal 0:
x-5=0

x=5

or

x+12=0

x=-12

However, since we are solving for the side length, the only possible answer is 5 (a shape can't have a side with a negative length.)

The Solution Is: 
\boxed{x=5}

I hope this helped!
-Benjamin

6 0
2 years ago
1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera
Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}

SP \approx 8.246

PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}

PA \approx  8.246

AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}

AZ \approx 8.246

ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}

ZS \approx 8.246

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

m_{SA} = \frac{6-(-4)}{6-(-4)}

m_{SA} = 1

m_{PZ} = \frac{4-(-2)}{-2-4}

m_{PZ} = -1

Since m_{SA}\cdot m_{PZ} = -1, diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)

M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)

M_{SA} = (-2,-2)+(3,3)

M_{SA} = (1,1)

M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)

M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)

M_{PZ} = (2,-1)+(-1,2)

M_{PZ} = (1,1)

Then, the diagonals SA and PZ bisect each other.

8 0
2 years ago
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