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3 years ago

Peyton says the sum of two numbers can never be zero. Explain why Peyton's reasoning is incorrect. Use an example.

Mathematics

1 answer:

Elena-2011 [213]3 years ago

4 0

Answer:

Well, there can be negative numbers added to their positive opposite.

Step-by-step explanation:

For example, -3 + 3 = 0

or you could 0 + 0 = 0

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Look at the figure below:

mr Goodwill [35]

Answer: A) ΔABD ≡ ΔECD

<u>Step-by-step explanation:</u>

Match up the letters of the congruent sides.

AD = ED and BD = CD

A D B D → ABD

E D C D → ECD

This can be written in any order as long as the letters are matched properly. A & E, D & D, B & C

ADB or BDA or ...

EDC CDE

The only correct option provided is option A.

6 0

4 years ago

Read 2 more answers

Solve by quadratic equation

Ymorist [56]

<h2>Question :</h2>

$\tt \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

<h2>Answer :</h2>

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

<h2>Explanation :</h2>

$\tt : \implies \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)(x+2) + (x-2)(x-2)}{(x-2)(x+2)} = \dfrac{5}{6}$

$\tt : \implies \dfrac{(x+2)^{2} + (x-2)^{2}}{(x-2)(x+2)} = \dfrac{5}{6}$

<u>Now, we know that</u> :

$\large \underline{\boxed{\bf{(a+b)^{2} = a^{2} + b^{2}+ 2ab}}}$
$\large \underline{\boxed{\bf{(a-b)^{2} = a^{2} + b^{2} - 2ab}}}$
$\large \underline{\boxed{\bf{(a+b)(a-b) = a^{2} - b^{2}}}}$

$\tt : \implies \dfrac{x^{2}+2^{2}+ 2 \times x \times 2 + x^{2}+2^{2} - 2 \times x \times 2 }{x^{2}-2^{2}} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2}+ 4 + \cancel{4x} + x^{2}+ 4 - \cancel{4x}}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{x^{2} + x^{2} + 4 + 4}{x^{2}-4} = \dfrac{5}{6}$

$\tt : \implies \dfrac{2x^{2} + 8}{x^{2}-4} = \dfrac{5}{6}$

<u>By cross multiply</u> :

$\tt : \implies (2x^{2} + 8)6= 5(x^{2}-4)$

$\tt : \implies 12x^{2} + 48 = 5x^{2}-20$

$\tt : \implies 12x^{2} + 48 - 5x^{2} + 20 = 0$

$\tt : \implies 7x^{2} + 68 = 0$

$\tt : \implies 7x^{2} + 0x + 68 = 0$

<u>Now, by comparing with ax² + bx + c = 0, we have</u> :

a = 7
b = 0
c = 68

<u>By using quadratic formula</u> :

$\large \underline{\boxed{\bf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}}}$

$\tt : \implies x = \dfrac{-(0) \pm \sqrt{(0)^{2} - 4(7)(68)}}{2(7)}$

$\tt : \implies x = \dfrac{0 \pm \sqrt{0 - 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{- 1904}}{14}$

$\tt : \implies x = \dfrac{\pm \sqrt{2\times 2\times 2\times 2\times 7\times 17}}{14}$

$\tt : \implies x = \dfrac{\pm \cancel{2} \times 2\sqrt{7\times 17}}{\cancel{14}}$

$\tt : \implies x = \dfrac{\pm2\sqrt{119}}{7}$

$\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}$

Hence value of $\bf x =\dfrac{\pm 2\sqrt{119}}{7}$

5 0

3 years ago

Read 2 more answers

Find the values of x and y. 3x - 5 5y - 4 y + 12

faltersainse [42]

3x - 5 | x = 1.6 repeating
5y - 4 | y = 0.8 or 4/5
y + 12 | y = -12
• I’m sorry if I didn’t understand it correctly

Have a nice day! :)

6 0

3 years ago

PLEASE ANSWER ASAP FOR BRAINLIEST

saul85 [17]

Answer:

Step-by-step explanation:

When you count the units it's 3, but because the scale factor is 4, you have to multiply it by that. 3 times 4 is 12.

3 0

4 years ago

Which of the m-values satisfy the following inequality?

dangina [55]

Answer:

a m=0

Step-by-step explanation:

5(0)+1 = 0+1 = 1

1<4

hope this helps

5 0

3 years ago