What is the area of trapezoid DEFG with coordinates D(2,3) E(4,3) F(6,1) G(2,1)?

2 answers:

8 0

The answer you are looking for is 6 square units I hope I am right sorry if I am wrong because I need to see the picture

Salsk061 [2.6K]3 years ago

4 0

Answer:

Answer is 6

Step-by-step explanation:

I took the quiz

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Solve the following triangle in which given sides are<br> a=7 b=7 c=9

blagie [28]

We have an isosceles triangle;
A=opposite angle side a.
B=opposite angle side b.
C=opposite angle side c.

A=B
Method 1:

We can divide the isosceles triangle in two right triangles,
hypotenuse=7
side=9/2=4.5

B=A=arccossine (4.5/7)=49.994799...º≈50º
C/2=90º-50º=40º ⇒ C=2*40º=80º

Answer:
a=7; A=50º
b=7; B=50º
<span>c=9; C=80º

Method 2:
Law of cosines:
a²=b²+c²-2bcCosA ⇒CosA=(a²-b²-c²)/(-2bc)
CosA=(49-49-81) / (-126)=0.642857
A=arco cos (81/126)≈50º

B=A=50º

A+B+C=180º
50º+50º+C=180º
C=180º-100º
C=80º

Answer:

</span>a=7; A=50º
b=7; B=50º
<span>c=9; C=80º</span>

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3 years ago

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cupoosta [38]

Answer:

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Step-by-step explanation:

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2 years ago

Multiply two and five-eighths negative two and three-fifths .

Yuki888 [10]

The answer is -6.825 in fraction form, it would be -6 33/40

4 0

3 years ago

Easy Sample Ap graphing 50pts! (Easy points!!) (ASAP) (URGENT) (Please check my other questions) Please

snow_lady [41]

Answer:

a) See attachment 1.

b) t² and D

c) See attachment 2.

d) g = 9.8 m/s² (1 d.p.)

Step-by-step explanation:

See attachment 1. The line of best fit is shown in red.

The quantities the student should graph in order to produce a <u>linear relationship</u> between the two quantities are t² and D.

Make a table of values of t² and D:

$\begin{array}{|l|l|l|l|l|l|}\cline{1-6} \rm t^2& 0.0196& 0.1024& 0.2116 & 0.3481& 0.3969\\\cline{1-6} \rm D & 0.10 & 0.50& 1.00 & 1.70 & 2.00\\\cline{1-6}\end{array}$

<u>Plot</u> a graph of D against t² and draw a line of best fit (see attachment 2).

From inspection of the graph, the line of best fit passes through the origin (0, 0) and (0.1024, 5.0). Therefore, use these two points to find the slope of the line:

$\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{0.5-0}{0.1024-0}=4.8828125...\:\rm m/s^2$