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oee [108]
3 years ago
7

The sides of a rectangle in the coordinate plane are parallel to the axes two of the vertices of the rectangle are (3,-2) and (-

4,-7) find coordinates for the other two verticals find the area of the rectangle

Mathematics
1 answer:
Llana [10]3 years ago
7 0

Answer:

C(-4,-2), D(3,-7)

Ar=35unit^{2}

Step-by-step explanation:

we procced to identify points:

A(-4,-7), B(3,-2)

for being the sides of a rectangle parallel to the axes, we find the other two vertices (VIEW GRAPH)

C(-4,-2), D(3,-7);  Area of the rectangle Ar=b*h, where

b=(X_{D}-X_{A}); h=(Y_{B}-Y_{D})

So Ar=(3-(-4))*(-2-(-7)) = (3+4)*(-2+7)=7*5=35unit^{2}

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You need to isolate the variable <em>h</em>. 

1) Subtract both sides by 2. 

-4h + 2 = 18
-4h + 2 - 2 = 18 - 2
-4h = 16

2) Divide both sides by -4 (the coefficient). 
(-4h) / -4 = 16 / -4
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If 1/2 gallon of paint covers 1/12 of a wall then how much paint is needed for the entire wall
netineya [11]

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3 years ago
what is the equation in slope intercept form of the line that crosses the x-axis at -2 and is perpendicular to the line represen
Verdich [7]

Answer:

y = -2x - 4

Step-by-step explanation:

So we have the x intercept and a line perpendicular to the one we are trying to find.

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the slope of the perpendicular line is 1/2 so that means the negative reciprocal is -2.

x intercept is when y=0, so the point is (-2, 0).  So that means the equation to make this point with the slope of -2 will look like this

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Now to get into the line we just plug in m and b into y = mx + b

y = mx + b

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4 years ago
If the original quantity is 20 and the new quantity is 16​, what is the percent​ decrease?
victus00 [196]

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Read 2 more answers
A train slows down as it rounds a sharp horizontal turn, going from 92.0 km/h to 54.0 km/h in the 16.0 s that it takes to round
Mnenie [13.5K]

Calculation of centripetal acceleration(ac):

a_c=\frac{v^2}{r}

we can change our speed in terms of m/s

v=54km/h=54*\frac{1000}{3600} m/s

now, we can plug these values

a_c=\frac{(54*\frac{1000}{3600})^2}{130}

a_c=1.73077m/s^2

Calculation of tangential acceleration(at):

a_t=\frac{\Delta v}{\Delta t}

now, we can plug values

a_t=\frac{(54-92)*\frac{1000}{3600} }{16}

now, we can simplify it

a_t=-0.65972 m/s^2

Calculation of resultant acceleration:

a=\sqrt{(a_c)^2+(a_t)^2}

now, we can plug values

a=\sqrt{(1.73077)^2+(-0.65972)^2}

a=1.85224m/s^2.............Answer

4 0
3 years ago
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