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Helen [10]
4 years ago
15

I need help finding out what shapes are Triangles or Quadrilaterals! Help me!

Mathematics
2 answers:
Margaret [11]4 years ago
7 0
Quadrilaterals have 4 sides, and triangles have 3 sides. So, you simply have to count the number of sides. 
ser-zykov [4K]4 years ago
4 0
Triangles are anything with three vertexes and three angles to it and are always closed.

Quadrilaterals are anything with four vertexes and four angles on them and are always closed.  Examples include a square, rectangle, trapezoid etc.
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A driver who drives at a speed of r mph for t hr will travel a distance d mi given by dequalsrt mi. How far will a driver travel
liubo4ka [24]

Answer:

At a speed of 57 mph for 8 hr a driver will travel 456 mi

Step-by-step explanation:

Here we have a summary of the letters for each variable:

Speed ---> r          (in units of mph)

Time ---> t            (in units of hr)

Distance ---> d     (in units of mi)

These three variables are related by the next formula:

d = rt

In the data they give to you: 57 mph and 8 hr, they are telling you the r and the t, respectively:

r = 57 mph

t = 8 hr

The only thing you have to do is replace the values:

d = rt    ---->     d = 57 mph x 8 hr

                        d = 456 mi

5 0
4 years ago
Colin invests £100 into his bank account.
kramer

The answer is $131.59

After year 1

104

Year 2

108.16

Year 3

112.49

Year 4

116.98

Year 5

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Year 7

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8 0
3 years ago
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Select the postulate of equality or inequality that is illustrated 5 = 5
Oliga [24]

reflexive postulate of equality

7 0
3 years ago
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Step-by-step explanation:

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3 0
4 years ago
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The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0
3 years ago
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