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quester [9]
3 years ago
8

30 POINTS. Please simplify this with all steps

0" id="TexFormula1" title="( \frac{5i^3t^4}{7})-2 " alt="( \frac{5i^3t^4}{7})-2 " align="absmiddle" class="latex-formula">
Mathematics
2 answers:
lesantik [10]3 years ago
8 0
Simplify the following:
(5 i^3 t^4)/7 - 2
i^3 = i^2×i = (-1) i = -i:
(5×-i t^4)/7 - 2
Put each term in (5 (-i) t^4)/7 - 2 over the common denominator 7:
(5 (-i) t^4)/7 - 2 = (-5 i t^4)/7 - 14/7:
(-5 i t^4)/7 - 14/7
(-5 i t^4)/7 - 14/7 = (-5 i t^4 - 14)/7:
Answer:  (-5 i t^4 - 14)/7
Brrunno [24]3 years ago
3 0
I^3= -1 
\frac{5i^{3}t^{4}  }{7} = -\frac{5it^4}{7}

\frac{5i^{3}t^{4} }{7} = -\frac{5it^4}{7} = answer-2- \frac{5t^4}{7}i
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Prove that a triangle with the sides a - 1 cm to root under a ​
vivado [14]

Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.  

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm  

Hence,

{AB}^2 = (a -1)^2 

Now We know that

(a- b)^2 = a^2+b^2 - 2ab

So;

{AB}^2= a^2 + 1^2 -2\times a \times1

{AB}^2 = a^2 + 1 -2a

Now;

{BC}^2 = (2\sqrt{a})^2= 4a

Also;

{CA}^2 = (a + 1)^2

Now We know that

(a+ b)^2 = a^2+b^2 + 2ab

{CA}^2= a^2 + 1^2 +2\times a \times1

{CA}^2 = a^2 + 1 +2a

{CA}^2 = AB^2 + BC^2

[By Pythagoras theorem]

a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a

Hence, {CA}^2 = AB^2 + BC^2

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

4 0
3 years ago
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