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luda_lava [24]
3 years ago
10

Need help please I will give brainlist

Mathematics
2 answers:
Vlad [161]3 years ago
6 0

Answer:

it is c

Step-by-step explanation:

c

alukav5142 [94]3 years ago
4 0

Answer:

Im pretty sure its c

Step-by-step explanation:

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What is 80/6 as a simplified fraction
Nat2105 [25]
Here is the answer for math problem

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write and equation of the line parallel to the line  2 x + 4 y = 20 2 x + 4 y = 20 and containing the point  ( −
MAXImum [283]

Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m represents the slope of the line

c represents the y intercept

The equation of the given line is

2x + 4y = 20

4y = - 2x + 20

Dividing through by 4, it becomes

y = - x/2 + 5

Comparing with the slope intercept form, slope = - 1/2

If two lines are parallel, it means that they have the same slope. Therefore, the slope of the line passing through (- 6, 3) is - 1/2

To determine the y intercept, we would substitute m = - 1/2, x = - 6 and y = 3 into y = mx + c. It becomes

3 = - 1/2 × - 6 + c

3 = 3 + c

c = 3 - 3 = 0

The equation becomes

y = - x/2

4 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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tekilochka [14]
Q1 of company A = 2.5
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Q1 of company B = 2
Q3 of company B = 5.5
Interquatile range = (5.5 - 2)/2 = 3.5/2 = 1.75

Therefore, t<span>he interquartile range for Company A employees is 2 more than the interquartile range for Company B employees.</span>
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Help me help me math question help me emerge
ioda

Answer:

Step-by-step explanation:

√13.7/(3.05^2+2.6)

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