Answer: It is not possible that two triangles that are similar and not congruent in spherical geometry.
Step-by-step explanation:
For instance, taking a circle on the sphere whose diameter is equal to the diameter of the sphere and inside is an equilateral triangle, because the sphere is perfect, if we draw a circle (longitudinal or latitudinal lines) to form a circle encompassing an equally shaped triangle at different points of the sphere will definately yield equal size.
in other words, triangles formed in a sphere must be congruent and also similar meaning having the same shape and must definately have the same size.
Therefore, it is not possible for two triangles in a sphere that are similar but not congruent.
Two triangles in sphere that are similar must be congruent.
Answer: 2.5
Step-by-step explanation:
Solve the numerator first.
We are left with:
To be able to work this out, we must convert the fraction to a decimal.
To do this, you can either go from a mixed number to an improper fraction and then from an improper fraction to a decimal, or you can divide the fractional part of the mixed number and add it to the whole number.
6+(1/4)
6+(0.25)
6+0.25
6.25
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Now, we know that this is a positive outcome because we have 2 negatives dividing.
6.25/2.5
To be able to work this out we have 1 more step to go. We have to convert these decimals to whole numbers; to do this, we multiply by 1 followed by as many zeros as the biggest number of decimals. In this case, the biggest number of decimals is 2, therefore; we must multiply by 1 followed by 2 zeros rather 100.
6.25*100/2.5*100
625/250
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Let's solve this.
625/250=2.5
500
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125'0
1250
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Answer:
9units²
Step-by-step explanation:
A triangle has 3 sides. Let the length of the three sides be a, b, c.
If the perimeter of the triangle is 18, then;
P = a+b+c
a+b+c = 18... 1
If the sum of the squares of the three side lengths is 128, this means;
a²+b²+c² = 128... 2
Since the triangle is also right angled, according to Pythagoras theorem;
c² = a²+b²... 3 (taking c as the hypotenuse)
Substitute equation 3 into 2;
(a²+b²)+c² = 128
c²+c² = 128
2c² = 128
c² = 128/2
c² = 64
c = √64
c = 8
Substituting c = 8 into 1
a+b+c = 18
a+b+8=18
a+b = 10 ...... 3
Substituting c = 8 into 2
a²+b²+c² = 128
a²+b²+8² = 128
a²+b²+64 = 128
a²+b²= 64..... 4
Solve equation 3 and 4 simultaneously;
a+b = 10
a²+b²= 64
From 4; a = 10-b
Substitute a = 10-b into 4;
(10-b)²+b² = 64
100-20b+b²+b² = 64
2b²-20b+100-64 = 0
2b²-20b+36 = 0
Divide through by 2:
b²-10b+18 = 0
b = 10±√10²-4(18)/2
b = 10±√100-72/2
b =( 10±√28)/2
b =(10±√4×7)/2
b = (10±2√7)/2
b = 5±√7
Hence b = 5+√7
If a+ b = 10
a+(5+√7 ) = 10
a = 10-(5+√7)
a = 10-5-√7
a = 5-√7
Area of the triangle A = 1/2 ab
A = 1/2 × (5+√7) × (5-√7)
A = 1/2 × (25-5√7+5√7-7)
A = 1/2 × (25-7)
A = 1/2 × 18
A = 9
Hence the are of the triangle is 9units²
To find the seventh term of the geometric sequence, divide first the fifth term by the fourth term to obtain the common ratio. -40 divide by -8 is 5. Then, multiply a5 by the common ratio, 5 twice or a4 by 5 three times. Both solutions will give, an a7 equal to -1000. Thus, a7 is -1000.
