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3 years ago

For all real numbers a and b, 2a • b = a2 + b2 True Or False ! Explain

Mathematics

2 answers:

Aliun [14]3 years ago

7 0

This is true only if one of a or b is negative
(a-b)^2 is 0
a^2+b^2-2ab=0
a^2+b^2=2ab

(A-b)^2 is 0
if a-b=0

cluponka [151]3 years ago

6 0

This may be equal with this formula of

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

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The function f(x) is represented by the table below. What are the corresponding values of g(x) for the transformation g(x) = 6f(

g100num [7]

Answer:

Given: The transformation g(x) =6f(x) ......[1]

To find the corresponding values of g(x) for the given transformation.

At x = -7, f(x) = 8

then,

By substituting the value of f(x) = 8 in [1] we have

$g(x) = 6 \cdot 8 =48$

Similarly,

For x= -3 , f(x) = 3, then

By substituting the value of f(x) = 3 in [1] we have

$g(x) = 6 \cdot 3 =18$

For x= 0, f(x) = -1 , then

By substituting the value of f(x) = -1 in [1] we have

$g(x) = 6 \cdot -1 =-6$

For x= 2, f(x) = 7, then

By substituting the value of f(x) = 7 in [1] we have

$g(x) = 6 \cdot 7=42$

For x= 10, f(x) = 5, then

By substituting the value of f(x) = 5 in [1] we have

$g(x) = 6 \cdot 5=30$

Therefore ; we have the corresponding values of g(x) as shown below;

x f(x) g(x)

-7 8 48

-3 3 18

0 -1 -6

2 7 42

10 5 30

5 0

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Read 2 more answers

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RSB [31]

True they could be congruent by ASA

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Judges in a diving competition gave scores of 9.3, 9 1/2, 9 4/5, 9.95 And 9 1/4. Which two scores were closest to one another?

Kazeer [188]

9.3 and $9\frac{1}{4}$ were closest to another.

Explanation

First we will convert each given fractions into decimal form. So....

$9\frac{1}{2}= 9+\frac{1}{2} = 9+0.50 =9.50\\ \\ 9\frac{4}{5}= 9+ \frac{4}{5}= 9+0.80=9.80\\ \\ 9\frac{1}{4}= 9+\frac{1}{4} =9+0.25= 9.25$

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That means , 9.3 and $9\frac{1}{4}$ were closest to another.

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BigorU [14]

B, a supplementary triangle because it all adds up to 180 degrees.

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ollegr [7]

Answer:

Divide both sides by -60. Note: when you divide an inequality by a negative number, switch the direction of the sign.

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