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Lera25 [3.4K]
2 years ago
11

(4 + 2) 2= Solve the equation

Mathematics
1 answer:
erica [24]2 years ago
3 0

uhh well 4 + 2 is 6- and it's in brackets so that's a start?

there's nothing between the brackets and the number

division: 3

multiplication: 12

addition: 8

subtraction: 4

expansion: 2(2+4) = 12

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Two staments that show the relationship between the values of the numbers 0.5 and 0.05
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<span> The place value relationship of 0.5 and 0.05 in the given number
=> 0.5 and 0.05 are decimal numbers because they are place next to the decimal points.
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3 years ago
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\bf \begin{array}{ll}&#10;n&a_n\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;6&50\\&#10;7&50+d\\&#10;8&50+d+d\\&#10;9&50+d+d+d\\&#10;10&50+d+d+d+d\\&#10;11&50+d+d+d+d+d\\&#10;&35&#10;\end{array}&#10;\\\\\\&#10;50+5d=35\implies 5d=-15\implies d=\cfrac{-15}{5}\implies d=-3&#10;\\\\\\&#10;\textit{we know d = -3, and we know }a_{11}=35\qquad thus

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\&#10;a_n=a_1+(n-1)d\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;d=\textit{common difference}\\&#10;----------\\&#10;d=-3\\&#10;n=11\\&#10;a_{11}=35&#10;\end{cases}&#10;\\\\\\&#10;a_{11}=a_1+(11-1)(-3)\implies 35=a_1+(11-1)(-3)&#10;\\\\\\&#10;35=a_1-30\implies 65=a_1

now we know what "d" is, and what a₁ is, so let's check the 30th term,

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\&#10;a_n=a_1+(n-1)d\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;d=\textit{common difference}\\&#10;----------\\&#10;d=-3\\&#10;n=30\\&#10;a_{1}=65&#10;\end{cases}&#10;\\\\\\&#10;a_{30}=65+(30-1)(-3)\implies a_{30}=65-87\implies a_{30}=-22
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