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2 years ago

Please help thank you in advance :)

Mathematics

1 answer:

marin [14]2 years ago

8 0

Answer:

1. 20

2. 10.

3. 2

4. 12

5.22

Step-by-step explanation:

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A polygon is shown:

Klio2033 [76]

Answer:

My brain just exploded. I tried imagining what the shape would look like, and that didn't work. Also, I tried drawing the shape and that didn't work. I have no idea what the shape looks like. If you still need help with this question, maybe you can repost it with an image attached so we can know what the shape looks like.

7 0

2 years ago

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<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

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What is the solution to the equation -3d/a^2-2d-8 + 3/d-4 = -2/ d+2

Alex73 [517]

Answer:

d=1

Step-by-step explanation:

$\frac{-3d}{d^2-2d-8} +\frac{3}{d-4} =\frac{-2}{d+2}$

Lets factor the denominator d^2 -2d-8

d^2 - 2d - 8 = (d-4)(d+2)

$\frac{-3d}{(d-4)(d+2)} +\frac{3}{d-4} =\frac{-2}{d+2}$

Now make the denominators same

LCD: (d-4)(d+2)

$\frac{-3d}{(d-4)(d+2)} +\frac{3(d+2)}{(d-4)(d+2)} =\frac{-2(d-4)}{(d+2)(d-4)}$

Denominators are same on both sides

So equate the numerators

-3d +3(d+2) = -2(d-4)

-3d +3d +6 = -2d +8

6 = -2d + 8

subtract 8 on both sides

-2 = -2d

So d=1

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2 years ago

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the point Z(3, -3) is rotated 180 degrees counterclockwise around the origin. what are the coordinates of the resulting point, Z

Leviafan [203]

Answer: -3 , 3

Step-by-step explanation:

thank me later;)

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3 years ago

Each cube in the prism is one cubic unit. <br> What is the volume of this rectangular prism?

Oksanka [162]

Answer:

I think the answer would be 120 cm3 .

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