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Trava [24]
3 years ago
15

Find the value of x in each case:

Mathematics
1 answer:
kipiarov [429]3 years ago
5 0
X = 69o
2x + 42 = 180
=> x = 69
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Translate the word phrase into a variable expression.
UNO [17]
The answer is b x/3-8
5 0
3 years ago
Pls help with 29..tank you
trapecia [35]

Answer:

See below

Step-by-step explanation:

Let the  2 roots be  A and A^2.

Then A^3 = c/a and A + A^2 = -b/a.

Using the identity a^3 b^3 = (a + b)^3 - 3ab^2 - 3a^2b:-

A^3 + (A^2)^3 =  ( A + A^2)^3 - 3 A.A^4 - 3 A^2. A^2

= (A + A^2)^3 - 3A*3( A + A^2)

Substituting:-

c/a + c^2/a^2 = (-b/a)^3 - 3 (c/a)(-b/a)

Multiply through by a^3:-

a^2c + ac^2 = -b^3 + 3abc

Factoring:-

ac(a + c) = 3abc - b^3

This is not the formula required in the question but let's see if the formula in the question reduces to this. If it does we have completed the proof.

a(c - b)^3 = a(c^3 - 3c^2b + 3cb^2 - b^3) = ac^3 - 3ac^2b + 3acb^2 - ab^3

c(a - b)^3 = c(a^3 - 3a^2b + 3ab^2 - b^3) = ca^3 - 3a^2bc + 3acb^2 - cb^3.

These are equal so we have

ca^3 - 3a^2bc + 3acb^2 - cb^3 = ac^3 - 3abc^2 + 3acb^2 - ab^3

The 3acb^2 cancel out so we have:-

a^3c - ac^3 =  3a^2bc - 3abc^2 + b^3c - ab^3

ac(a^2 - c^2) = 3abc( a - c) + b^3(c - a)

ac(a + c)(a -c) = 3abc(a - c) - b^3 (a - c)

Divide through by (a - c):-

ac(a + c) = 3abc - b^3 , which is the result we got earlier.

This completes the proof.

 

3 0
3 years ago
Read 2 more answers
Can someone help please?
Over [174]

Answer:

its A

Step-by-step explanation:


8 0
3 years ago
Staci has 7 more cats than Taylor. Write an expression that illustrates how many cats Staci has. use x to represent the number o
jasenka [17]
Taylor=T   Staci=S
T=x   S=x+7
3 0
3 years ago
Check whether these statements are wff or not:(a) (p˅q) ∧∼r
Free_Kalibri [48]

Answer:

It is a well formed formula

Step-by-step explanation:

1 - p,q,r are well formed formulas.

2 - p \ \lor \ q  is a well formed formula as well.

3 - \neg r is a well formula as well

4 - (\ p \ \lor \ q) \ \land \ \neg r is a well formula as well.

4 0
3 years ago
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