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vladimir2022 [97]

4 years ago

13

3) What is the reciprocal of 4? (a))-4 b) 0 C 1/4 D 4

2 answers:

nadezda [96]4 years ago

7 0

Answer:

C) 1/4

Step-by-step explanation:

ryzh [129]4 years ago

3 0

Answer:

The reciprocal of 4 is 1/4

Step-by-step explanation:

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Leah gets paid $795 biweekly. What is the recommended monthly rent that she can afford?

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4 years ago

The 4th and the last terms of an A.P. are 11 and 89 respectively. If there are 30 terms in the A.P., find the A.P. and its 23rd

Natali5045456 [20]

$\underline{\underline{\large\bf{Given:-}}}$

$\red{\leadsto}\:$ $\textsf{}$ $\sf Number \: of \:terms \: in \: A.P,n = 30$

$\red{\leadsto}\:$ $\textsf{}$ $\sf Fourth \: term ,a_4 = 11$

$\red{\leadsto}\:$ $\textsf{}$ $\sf last\:term, a_{30} = 89$

$\underline{\underline{\large\bf{To Find:-}}}$

$\orange{\leadsto}\:$ $\textsf{ }$ $\sf The \: A.P.$

$\orange{\leadsto}\:$ $\textsf{ }$ $\sf 23rd\: term, a_{23}$

$\\$

$\underline{\underline{\large\bf{Solution:-}}}\\$

The nth term of A.P is determined by the formula-

$\green{ \underline { \boxed{ \sf{a_n = a+(n-1)d}}}}$

where

$\sf a = first \:term$
$\sf a_n = nth \: term$
$\sf n = number \:of \:terms$
$\sf d = common \: difference$

Since ,

$\sf a_4 = 11$

$\longrightarrow$ $\sf a+(4-1) d= 11$

$\longrightarrow$ $\sf a+3d= 11\_\_\_(1)$

$\sf a_{30}= 89$

$\longrightarrow$ $\sf a+(30-1)d=89$

$\longrightarrow$ $\sf a+29d= 89\_\_\_(2)$

<u>Subtracting equation (1) from equation(2)</u>

$\begin{gathered}\\\implies\quad \sf a+29d-(a+3d) = 89-11 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a+29d-a-3d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a-a+29d-3d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf 26d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf d = \frac{78}{26} \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf d = 3 \\\end{gathered}$

Putting the value of d in equation (1) -

$\begin{gathered}\\\implies\quad \sf a+3(3) = 11 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a = 11-9 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a = 2 \\\end{gathered}$

First term of A.P, a = 2

Second term of A.P., $\sf a_2= 2+(2-1)\times 3$

$\quad\quad\quad\sf =2+3$

$\quad\quad\quad\sf =5$

Third term of A.P., $\sf a_3= 2+(3-1)\times 3$

$\quad\quad\quad\sf =2+6$

$\quad\quad\quad\sf =8$

$\longrightarrow$ Thus , The A.P is 2,5,8,. . . . . .

<u>Now,</u>

$\begin{gathered}\\\implies\quad \sf a_n = a+(n-1)d \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23 }= 2+(23-1)\times 3 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 2+22 \times 3 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 2+66 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 68 \\\end{gathered}$

$\longrightarrow$ Thus , 23rd term is 68.

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3 years ago

Round 974657 to the nearest ten thousand

yuradex [85]

Answer:

970000

Step-by-step explanation:

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4 years ago