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3 years ago

7

Natalie works in a department store selling clothing, She makes a guaranteed salary

1 answer:

marusya05 [52]3 years ago

5 0

Answer:

215

Step-by-step explanation:

subtract 30 from 225

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Solve x-5y=-30, 3x+5y=10 show your work

marin [14]

1) x = 5y - 30

2) x = - 1.6y + 3.3

<em>im assuming these are 2 different questions</em>

<h3>Explanation:</h3><h3>1)</h3>

Start with the equation:

x - 5y= -30

add 5y to both sides:

x = 5y - 30

Your answer is x = 5y - 30

<h3>2)</h3>

Start with the equation:

3x+5y=10

subtract 5y from both sides:

3x = -5y + 10

divide by 3 to get x singular

x = - 1.6y + 3.3

Your answer is x = - 1.6y + 3.3

Hope this helps :)

6 0

3 years ago

Adam had some candy to give to his three children. He first took seven pieces for himself and then evenly divided the rest among

Katen [24]

Adam took 7 pieces so subtract it from the total.
$13 - 7 = 6$
He divided the rest to his 3 children
$\frac{6}{3} = 2$
Each child recieved 2 candies.

8 0

4 years ago

Read 2 more answers

Factorize 16x square-4x-4x+1

marin [14]

${16x}^{2} - 4x - 4x + 1 \\ = 4x(4x - 1) - 1(4x - 1) \\ = (4x - 1)(4x -1 )$

4 0

3 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.

3 0

3 years ago

mr_godi [17]

Answer : 47 miles per hour
50 mph = 50*5280 ft / hour = 50*5280 / 3600 ft / sec 73.33 ft/sec so Morrius speed is 70.433 ft/sec in mph this is (70.33* 3600 )/ 5280 = 47 mph

6 0

3 years ago