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Luden [163]
3 years ago
11

Please help me , i have no idea what the answer is

Mathematics
1 answer:
liraira [26]3 years ago
4 0

So,

We see in Statement 3 that WY = WY.  This seems obvious, but it ensures that quantities don't change their values.  We call this property of something being equal to itself, the "reflexive property."

The correct answer is the third option.

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What is the value of the discriminant, b2 − 4ac, for the quadratic equation 0 = x2 − 4x + 5, and what does it mean about the num
forsale [732]
(-4)^2-4(1)(5)
16-4(1)(5)
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16-20
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The discriminant is negative so there are no real number solutions.
3 0
3 years ago
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What is the slope of a line parallel to the line whose equation is 2. – y= -7. Fully
g100num [7]

Answer:

Step-by-step explanation:

Hmmmmm you did not give the full equation for the original line.

However, parallel lines have identical slopes. Whatever coefficient (m) is associated with the x value, will be the slope of the parallel line when the equation is put in the form

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7 0
2 years ago
Helpppp pleaseeeeeeeeeeeee
astraxan [27]

Answer:

$1.30

Step-by-step explanation:

91/14/5

7 0
3 years ago
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If the are of a square is multiplied by 9, the area becomes 16 square inches. Find the length of the X side of the square.
Volgvan
X^2 = 16/9
X = 4/3

Length of X is 1 1/3   inches
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%20%5Csf%20%5Chuge%7B%20question%20%5Chookleftarrow%7D" id="TexFormula1" title=" \sf \huge
BabaBlast [244]

\underline{\bf{Given \:equation:-}}

\\ \sf{:}\dashrightarrow ax^2+by+c=0

\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.

\sf We\:know,

\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}

\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}

\underline{\large{\bf Identities\:used:-}}

\boxed{\sf (a+b)^2=a^2+2ab+b^2}

\boxed{\sf (√a)^2=a}

\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}

\boxed{\sf \sqrt{\sqrt{a}}=a}

\underline{\bf Final\: Solution:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}

\bull\sf Apply\: Squares

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}

\bull\sf Put\:values

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}

\bull\sf Simplify

\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}

\underline{\bf More\: simplification:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}

\underline{\Large{\bf Simplified\: Answer:-}}

\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}

5 0
2 years ago
Read 2 more answers
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