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vova2212 [387]
4 years ago
9

Solve: 5y+34 = -2(-7y+1)

Mathematics
2 answers:
Blizzard [7]4 years ago
6 0

Answer:

Slope = 1.200/2.000 = 0.600

 x-intercept = 34/3 = 11.33333

 y-intercept = 34/-5 = -6.80000

Step-by-step explanation:

1.1     Solve   3x-5y-34  = 0

Tiger recognizes that we have here an equation of a straight line. Such an equation is usually written y=mx+b ("y=mx+c" in the UK).

"y=mx+b" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis.

In this formula :

y tells us how far up the line goes

x tells us how far along

m is the Slope or Gradient i.e. how steep the line is

b is the Y-intercept i.e. where the line crosses the Y axis

The X and Y intercepts and the Slope are called the line properties. We shall now graph the line  3x-5y-34  = 0 and calculate its properties

Geometric figure: Straight Line

 Slope = 1.200/2.000 = 0.600

 x-intercept = 34/3 = 11.33333

 y-intercept = 34/-5 = -6.80000

Sergeu [11.5K]4 years ago
4 0

Answer:

y=1.8

Step-by-step explanation:

5y+34=-2(-7y+1)

5y+34=14y-2

34=19y-2

36=21y

1.7

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Answer:2

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3 years ago
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cube m holds a maximum of 64 unit cubes. each unit cube has an edge length of 1/2 centimeter. what is the length of cube m?
yulyashka [42]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
A regulation soccer field for international play is a rectangle with a length between 100 m and I 1 O m and a width between 64 1
Natalija [7]

Answer:

  • The smallest area the field could be is 6,400 m²
  • The largest area the field could be is 8,250 m²

Step-by-step explanation:

Given;

smallest possible length of the international soccer field, L₀ = 100 m

smallest possible breadth of the international soccer field, B₀ = 64 m

Largest possible length of the international soccer field, L₁ = 110 m

Largest possible breadth of the international soccer field, B₁ = 75 m

Area of a rectangle is given by;

A = L x B

The smallest area the field could be is calculated as;

A₀ = L₀ x B₀

A₀ = 100 m x 64 m

A₀ = 6,400 m²

The largest area the field could be is calculated as;

A₁ = L₁ x B₁

A₁ = 110 m x 75 m

A₁ = 8,250 m²

6 0
3 years ago
What is the following coordinates of (2,0) and (0,4)
34kurt

Answer:

(2, 0) and (0, 4) are coordinates in the form of (x, y)

(2, 0) -> x = 2, y = 0

(0, 4) -> x = 0, y = 4

8 0
3 years ago
What is 5.316 - 1.942 (show ur work)
Fiesta28 [93]

Answer:

3.374

Step-by-step explanation:

\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}

\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}

\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4

\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}

\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}

\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}

\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Borrow\:}1\mathrm{\:from\:}5\mathrm{.\:\:The\:remainder\:is\:}4

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}

\mathrm{Add\:}1\mathrm{\:ten\:to\:}3:\quad \:10+3=13

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Borrow\:}1\mathrm{\:from\:}13\mathrm{.\:\:The\:remainder\:is\:}12

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:11-4=7

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}

\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}

\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-1=3

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}

=3.374

Hence the correct answer is 3.374

7 0
2 years ago
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