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tankabanditka [31]
3 years ago
12

Good morning, can someone help me solve this problem? I already did some of it but I need help, thank you

Mathematics
1 answer:
UNO [17]3 years ago
7 0

Answer:   (-4, -1)

This means x = -4 and y = -1 pair up together.

======================================================

You're on the right track. You'll replace the 'x' with 3y-1, but you made a typo. I think you meant to write the subtraction symbol instead of the equal sign.

Here's what your steps could look like

-2x + 5y = 3

-2(3y-1) + 5y = 3

-6y + 2 + 5y = 3

-y + 2 = 3

-y = 3-2

-y = 1

y = -1

x = 3y-1

x = 3(-1) - 1

x = -3-1

x = -4

Therefore, (x, y) = (-4, -1) is the solution

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

As the steps above show, I replaced each copy of x in the second equation with (3y-1). This is due to x = 3y-1. From there, we isolate y. Once you determine that y = -1, it's plugged into the first equation to find x.

If you were to graph x = 3y-1 and -2x+5y = 3 on the same xy coordinate grid, then the two lines would intersect at (-4, -1)

Also, note how plugging x = -4 and y = -1 into the second equation leads to...

-2x+5y = 3

-2(-4)+5(-1) = 3

8 - 5 = 3

3 = 3

We get the same value on both sides, so that verifies the second equation. I'll let you verify the first equation.

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The area of an 14-cm-wide rectangle is 322 cm2. What is its length?<br> The length is<br> cm.
Anika [276]

Answer:

The length of rectangle is 23 cm.

Step-by-step explanation:

<u>DIAGRAM</u> :

\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large{14\ cm}}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large{14\ cm}}\put(-0.5,-0.4){\bf}\put(-0.5,3.2){\bf}\put(5.3,-0.4){\bf}\put(5.3,3.2){\bf}\end{picture}

\begin{gathered}\end{gathered}

<u>SOLUTION</u> :

Here's the required formula to find the length of rectangle :

{\longrightarrow{\pmb{\sf{A_{(Rectangle)}  = l \times b}}}}

  • A = Area
  • l = length
  • b = breadth

Substituting all the given values in the formula to find the length of rectangle :

\begin{gathered}\qquad{\longrightarrow{\sf{A_{(Rectangle)}  = l \times b}}}\\\\\qquad{\longrightarrow{\sf{322 = l \times 14}}}\\\\\qquad{\longrightarrow{\sf{322 = 14l}}}\\\\\qquad{\longrightarrow{\sf{l =  \dfrac{322}{14}}}}\\\\\qquad{\longrightarrow{\sf{l =  \cancel{\dfrac{322}{14}}}}}\\\\\qquad{\longrightarrow{\sf{l = 23 \: cm}}}\\\\\qquad{\star{\underline{\boxed{\sf{ \pink{l = 23 \: cm}}}}}}\end{gathered}

Hence, the length of rectangle is 23 cm.

\begin{gathered}\end{gathered}

<u>LEARN</u><u> </u><u>MORE</u> :

\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}

\rule{300}{2.5}

6 0
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Answer:

<h2>The bus was 1250 meters away from school</h2>

Step-by-step explanation:

Given that the total distance of the school from home is 6km

converting 6km to meters we have 6,000m

Since the bus stopped at 4750 meters from his home.

The remaining distance is the distance the bus is from his school

i.e 6000-4750= 1250 meters

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A rock breaks loose from a cliff and plunges towards the ground 400 feet below. The distance 'd' that the rock falls in 't' seco
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You are the treasurer for a local charity. At the beginning of the month the balance was $820.64. The charity received donations of $500.00, $55.00 and $25.00. You have an electric bill of $40.64, bill for postage of $12.75, and rent for $445.00.
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Santa's elves are selling
Andreas93 [3]

cost of one peppermint cookies = $ 0.6

cost of one cinnamon sugar cookies = $ 0.3

<h3><u>Solution:</u></h3>

Let "p" be the cost of one peppermint cookies

Let "c" be the cost of one cinnamon sugar cookies

<u><em>To find: cost of each cookie</em></u>

<h3><u><em>On  the first day, they sold 120  peppermint cookies and 30  cinnamon sugar cookies for a  total of $81</em></u></h3>

We can frame a equation as:

120  peppermint cookies x cost of one peppermint cookies + 30  cinnamon sugar cookies x cost of one cinnamon sugar cookies = $ 81

120 \times p + 30 \times c = 81

120p + 30c = 81  --------- eqn 1

<h3><u><em>The next day  they made $60 by selling 70  peppermint cookies and 60  cinnamon sugar cookies</em></u></h3>

70  peppermint cookies x cost of one peppermint cookies + 60  cinnamon sugar cookies x cost of one cinnamon sugar cookies = $ 60

70 \times p + 60 \times c = 60

70p + 60c = 60 --------- eqn 2

Let us solve eqn 1 and eqn 2 to find values of "p" and "c"

Multiply eqn 1 by 2

240p + 60c = 162  --- eqn 3

Subtract eqn 2 from eqn 3

240p + 60c = 162

70p + 60c = 60

(-) -------------------------

170p = 102

<h3>p = 0.6</h3>

Substitute p = 0.6 in eqn 1

120p + 30c = 81

120(0.6) + 30c = 81

72 + 30c = 81

30c = 9

<h3>c = 0.3</h3>

<u><em>Summarizing the results:</em></u>

<em>cost of one peppermint cookies = $ 0.6</em>

<em>cost of one cinnamon sugar cookies = $ 0.3</em>

6 0
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the amount of men in recital is 3 because it says men first in the answer when doing ratios always first word is the first number


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