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3 years ago

Pls help thanks plsssssasa

Mathematics

2 answers:

maw [93]3 years ago

8 0

The distributive property

Korvikt [17]3 years ago

6 0

It would would be the distributive property! Hope that this helped!

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The value of y varies directly as x, y is 9 when x is -3 what is the constant of variation k

exis [7]

Direct variation has the form y=kx. Given a point we can solve for the constant k.

y=kx, when y=9 and x=-3 becomes:

9=-3k

k=-3 so the constant of variation is negative 3

7 0

3 years ago

Solve for x ~<br><br><img src="https://tex.z-dn.net/?f=%5Cbold%5Cpink%7Bx%20%7B%7D%5E%7B2%7D%20%2B%205x%20-%206%20%3D%200%7D" id

Troyanec [42]

$\qquad \qquad \rm \large \dag \: Answer \: \dag$

We have been given a Quadratic equation, where we have to find the real values of x ~

So, let's proceed with middle term splitting method :

$\qquad \sf \dashrightarrow \: {x}^{2} + 5x - 6 = 0$

$\qquad \sf \dashrightarrow \: x {}^{2} + 6x - x - 6 = 0$

$\qquad \sf \dashrightarrow \: {x(}^{} x + 6) - 1(x + 6) = 0$

$\qquad \sf \dashrightarrow \: (x + 6)(x - 1) = 0$

Now, we have two conditions here~

Either (x + 6) = 0

$\qquad \sf \dashrightarrow \: x = - 6$

or (x - 1) = 0

$\qquad \sf \dashrightarrow \: x = 1$

4 0

2 years ago

Three points on the graph of the function f(x)f(x) are \{(0, 4), (1, 6), (2, 8)\}{(0,4),(1,6),(2,8)}. Which equation represents

Blababa [14]

Answer:

y = 2x + 4

Step-by-step explanation:

Slope: (6-4)/(1-0) = 2

y = 2x + 4

7 0

3 years ago

Help me out? Thank you very much :)

astra-53 [7]

12.56 = 2(3.14) square root (L/32)
12.56 = 6.28 square root (L/32)
square root (L/32) = 2
L/ 32 = 2^2
L/ 32 = 4
L = 4 (32)
L = 128

5 0

4 years ago

Read 2 more answers

Pls help me.<br>proove it ^

aivan3 [116]

Answer:

We verified that $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

Hence proved

Step-by-step explanation:

Given equation is $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

We have to prove that $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

That is to prove that LHS=RHS

Now taking RHS

$\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

$=\frac{a+b+c}{2}[a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2]$ (using $(a-b)^2=a^2-2ab+b^2$ )

$=\frac{a+b+c}{2}[2a^2-2ab+2b^2-2bc+2c^2-2ac]$ (adding the like terms)

$=\frac{a+b+c}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ac]$

$=\frac{a+b+c}{2}\times 2[a^2+b^2+c^2-ab-bc-ac]$

$=a+b+c[a^2+b^2+c^2-ab-bc-ac]$

Now multiply the each term to another each term in the factor

$=a^3+ab^2+ac^2-a^2b-abc-a^2c+ba62+b^3+bc^2-ab^2-b^2c-abc+ca^2+cb^2+c^3-abc-bc^2-ac^2]$

$=a^3+b^3+c^3-3abc$ (adding the like terms and other terms getting cancelled)

$=a^3+b^3+c^3-3abc$ =LHS

Therefore LHS=RHS

Therefore $a^3+b^3+c^3-3abc=\frac{a+b+c}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$

Hence proved.

8 0

4 years ago