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Harman [31]
3 years ago
11

The question​ of the week! (Please show your work!)

Mathematics
1 answer:
solniwko [45]3 years ago
5 0

3 -  \frac{x}{8}  = 1 \frac{1}{4} \\ 3 - 1 \frac{2}{8}  =  \frac{x}{8}  \\  \frac{24}{8}  -  \frac{10}{8}  =  \frac{x}{8}  \\  \frac{14}{8}  =  \frac{x}{8}  \\ 14 = x
14 pieces that are 1/8th a yard each
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I NEED ANSWERS FAST PLEASE ILL GIVE BRAINLIEST
malfutka [58]

Answer:

4. C

5. A

6. D

Step-by-step explanation:

3 0
3 years ago
(57x2:3)+32x450-[39-(4+2):2[+8=
tigry1 [53]

Answer:

57×2/3+32×450-[39-4+2]/2+8

57×2/3+32×450-(39-6/2)+8

57×2/3+32×450-36+8

57×2/3+14400-28

57×2/3+14372

38+14372

14410

8 0
2 years ago
What is greater 3.6 or 0.94
nasty-shy [4]
3.6 > 0.94 Hope this helps.
6 0
3 years ago
Read 2 more answers
<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
The table shows the distance in feet, y, that a rocket traveled in x seconds.
Murrr4er [49]

Answer:

The correct option is D. y = 25x^{2}.

Step-by-step explanation:

i) From the table and our own observation and a trial and error approach we

  can clearly see that the equation that matches y in feet ( the distance

  traveled by the rocket) and x in seconds ( the time elapsed) is given by the

  equation y = 25x^{2}

 Therefore the correct option is D. y = 25x^{2}.

7 0
3 years ago
Read 2 more answers
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