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3 years ago

(c)x(c)x(c)x(c)x(c)x(c)?Help please

Mathematics

2 answers:

vladimir2022 [97]3 years ago

7 0

Answer:

.........................................

Step-by-step explanation:

Furkat [3]3 years ago

7 0

C(x)^5c?
Am I suppose to simplify...
Not sure what the question

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Which is more 2/3 or 2/4

goblinko [34]

The answer would be 2/3 is greater than 2/4. Two- thirds is equivalent to 0.67 while 2/4 is equivalent to 1/2 or 0.5. Of course, 0.67 is greater than 0.5.

6 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]

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

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

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

$\sf We\:know,$

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

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

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

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

$\boxed{\sf (√a)^2=a}$

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

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

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

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

$\bull\sf Apply\: Squares$

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

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

$\bull\sf Put\:values$

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

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

$\bull\sf Simplify$

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

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

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

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

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

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

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2 years ago

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Help please, I really need to get this test passed

ad-work [718]

Answer:

the secound ansewr

Step-by-step explanation:ik

3 0

3 years ago

:) PLEASE!!!! HELP! some food while you solve <3

Dominik [7]

Answer:

1/3

Step-by-step explanation:

rise over run. Find two places that land on a line I picked (0,0) and (3,1) then count up one from the first point and over 3 for the second.

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3 years ago

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Can you solve this problem please

Afina-wow [57]

Answer:

k = 11

Step-by-step explanation:

Given the points are collinear then the slopes between consecutive points are equal.

Using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (5, 1) and (x₂, y₂ ) = (1, - 1)

m = $\frac{-1-1}{1-5}$ = $\frac{-2}{-4}$ = $\frac{1}{2}$

Repeat with another 2 points and equate to $\frac{1}{2}$

with (x₁, y₁ ) = (1, - 1) and (x₂, y₂ ) = (k, 4)

m = $\frac{4+1}{k-1}$ , then

$\frac{5}{k-1}$ = $\frac{1}{2}$ ( cross- multiply )

k - 1 = 10 ( add 1 to both sides )

k = 11

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3 years ago

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