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Butoxors [25]
3 years ago
13

What is 5 (3x + 2y) -2 (3x + 2y) ?

Mathematics
1 answer:
Law Incorporation [45]3 years ago
4 0

Answer:

9x+6y is your simplified answer

Step-by-step explanation:

Your welcome

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If an equation has an identity how many equations does it have
melamori03 [73]

Answer:


Is there anyway you can explain it more?


8 0
3 years ago
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Pls help me on this question
Step2247 [10]

Answer:H

Step-by-step explanations

Hope this helpes

4 0
3 years ago
Maria’s family hosts a barbeque every month in the summer. In May, they bought 3.2 pounds of hamburger and 3 packages of hot dog
Sindrei [870]
Given:

May:
hamburger: 3.2 pounds * 4.25 = 13.60
hotdogs:     3 packages * 4      = 12
Total cost: 13.60 + 12 = 25.60

June:
hamburger: 3.2 pounds * 1 1/2 = 4.8 pounds * 4.25 = 20.4
hotdogs:      3 packages * 1/3    = 1 package * 4        =   4
Total cost: 20.4 + 4 = 24.4

25.60 - 24.40 = 1.20

They spent more on May by $1.20. Choice C.
5 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
In an international film festival, a penal of 11 judges is formed to judge the best film. At last two films FA and FB were consi
valentina_108 [34]

Answer:

The answer is "0.4329 ".

Step-by-step explanation:

P( three in favor of FA)

Select 3 out of 6 FA supporters then select 2 out of 5 FB supportive judges  

=\frac{^{6}_{C_{3}}\times ^{5}_{C_{2}}}{^{11}_{C_{5}}}\\\\=\frac{\frac{6!}{3!(6-3)!}\times \frac{5!}{2!(5-2)!}}{\frac{11!}{5!(11-5)!}}\\\\=\frac{\frac{6!}{3! \times 3!}\times \frac{5!}{2! \times 3!}}{\frac{11!}{5! \times 6!}}\\\\=\frac{\frac{6 \times 5 \times 4 \times 3!}{3 \times 2  \times 1\times 3!}\times \frac{5  \times 4  \times 3!}{2 \times 1 \times 3!}}{\frac{11  \times10  \times 9  \times 8 \times 7  \times 6! }{5 \times 4 \times 3  \times 2  \times 1 \times 6!}}\\\\

=\frac{ (5 \times 4) \times(5  \times 2)}{(11  \times 3  \times 2 \times 7 )}\\\\=\frac{ 20 \times 10 }{(11 \times 42)}\\\\=\frac{ 200 }{462}\\\\=\frac{100 }{231}\\\\=0.4329

6 0
3 years ago
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