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Setler [38]
1 year ago
10

What is the solution to the system of equations? y = A system of equations. Y equals StartFraction 2 over 3 EndFraction x plus 3

. X equals negative 2.X + 3 x = –2
Mathematics
2 answers:
Sauron [17]1 year ago
5 0

Answer:

The solution to the given system of equations is (-2,\frac{5}{3})

Therefore the values of x and y are x=-2 and y=\frac{5}{3}

Step-by-step explanation:

Given equations can be written as

y=\frac{2}{3}x+3\hfill (1)

x=-2x+3x=-2\hfill (2)

Solving equation(2) we get

x=-2

Substitute x=-2 in equation (1) we get

y=\frac{2}{3}(-2)+3

=-\frac{4}{3}+3

=\frac{-4+9}{3}

=\frac{5}{3}

Therefore the values of x and y are  x=-2 and y=\frac{5}{3}

The solution to the given system of equations is (-2,\frac{5}{3})

eimsori [14]1 year ago
3 0

Answer:

It’s B and E

Step-by-step explanation:

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Bella has joined a new gym in town. The cost of membership is $25 per month. The after-hours policy at the gym allows her to wor
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The cost of membership of a month = $25

Let 'n' be extra the number of hours Bella worked on.

The cost for working on extra hours = $4

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(Monthly cost of membership) + ( cost for extra hours \times number of hours extra worked on )  = Total monthly bill received

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Titus is asked to prove hexagon FEDCBA is congruent to hexagon
Goshia [24]

Answer:

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Step-by-step explanation:

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5 0
1 year ago
The length of the shadow of a pole on level ground increases by 90m when the angle of elevation of the sun changes from 58° to 3
viva [34]

Answer:

119.45 m

Step-by-step explanation:

Given:

When angle of elevation of the sun changes from 58° to 36° the length of shadow of a pole increases by 90 m.

To find:

Length of pole = ?

Solution:

Kindly refer to the attached image.

\triangle ABC represents the 1st angle of elevation of sun i.e.  58°

\triangle ABD represents the 2nd angle of elevation of sun i.e.  36°

Change in shadow is represented by CD = 90 m

Let height of pole, AB = h m

Let side BC = x m

Now, let us apply tangent rules in \triangle ABC, \triangle ABD one by one:

tan\theta = \dfrac{Perpendicular}{Base}\\\Rightarrow tan58^\circ=\dfrac{AB}{BC}\\\Rightarrow tan58^\circ=\dfrac{h}{x}\\\Rightarrow x = 0.624h ..... (1)

tan36^\circ = \dfrac{h}{x+90}

Putting value of x using equation (1):

tan36^\circ = \dfrac{h}{0.624h+90}\\\Rightarrow 0.726\times 0.624h+0.726\times 90 = h\\\Rightarrow h-0.453h =65.34\\\Rightarrow \bold{h = 119.45\ m}

119.45 m is the height of pole.

7 0
2 years ago
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