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2 years ago

9.45 = 28.2 can you answer

Mathematics

2 answers:

mixas84 [53]2 years ago

8 0

Answer:

9.45g= 28.2

Divide through by 9.45

g=28.2/9.45

g=2.98 to 2s.f

Mrrafil [7]2 years ago

4 0

As the answer is 9.45g=28.2, we can figure out easily that g is just 28.2 divide by 9.4. or 28.2/9.4

Dividing this we can get 3 easily, this is simple division.

If you have any question just ask in the comments.

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Using the Triangle Sum Theorem, find x and the missing angles (2x +1) (5x +5) x= Angle (2x+1) = Angle (5x+5) =

Arte-miy333 [17]

Answer:

x = 12

(2x + 1)° = 25°

(5x + 5)° = 65°

Step-by-step explanation:

By the property of a triangle,

Sum of all angles of a triangle is 180°

In the given right triangle,

(2x + 1)° + 90° + (5x + 5)°= 180°

7x + 96 = 180

7x = 180 - 96

x = $\frac{84}{7}$

x = 12

Angle (2x + 1)° = 2(12) + 1 = 25°

Angle (5x + 5)° = 5(12) + 5 = 65°

Therefore, all three angles of the right triangle are 25°, 90° and 65°.

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2 years ago

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2 years ago

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

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3 years ago