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NikAS [45]
2 years ago
15

Please help me I don't know what this is I will mark brainliest

Mathematics
1 answer:
lyudmila [28]2 years ago
7 0

Answer:

improper fractions are mixed numbers that can be multiplied by bottom to top like 2/4 x 4/5 so is 2 and 4 from 2/4 and 4/5 you can also do bottom to top like 2 x 5=10 and 4 x 4= 16 now you can add 10 + 16=26

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A. 30 -1 =7<br> B.<br> 32 = 8<br> 1) How can we get Equation B from Equation A?
mixer [17]

Answer:

you +1 on each side

Step-by-step explanation:

a is 30-1=7

b is 32=8

to get from "a" to "b" you simply add one on both sides

32 - 1 = 7

+ 1 +1

------------------

32=8

6 0
3 years ago
PLEASE HELP!!!!!! DUE TODAY!!!!!! PLEASEEEEEEEE!!!!!!!!
Sergio039 [100]
Once again it's (2a,0) I took this and got it right
4 0
4 years ago
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Aaron and Maria collected shells one morning at the beach. Maria had 3 times as many as Aaron, but they would have had the same
Hoochie [10]
Let the amount of shells collected by Aaron be x and the amount of shells collected by Maria be y respectively. Then the following system of equations should be solved:
\left \{ {{3x=y} \atop {x+8=y-6}} \right.

First of all, express y out of first equation:
y=3x

Then, replace y in second equation with found expression and solve it:
x+8=3x-6
2x=14
x=14/2=7

Finally, replace x in first equation with found value and solve it too:
3*7=y
y=21

So, Aaron has found 7 shells while Maria has found 21.


7 0
3 years ago
Which equation best represents the relationship between x and y in the graph
Naily [24]
Is A y=2x-6 I think!!!
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}}}}

5 0
2 years ago
Read 2 more answers
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