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3 years ago

I don’t understand what exactly they are looking for. I am completely stumped on #24

Mathematics

1 answer:

Mrac [35]3 years ago

7 0

Answer:

Step-by-step explanation:

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The graph shown is the graph of which function?

alukav5142 [94]

Where the pictures of the graph

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4 years ago

Here is a linear equation: y = 1/4x +5/4

docker41 [41]

Step-by-step explanation:

(1,1.5) is the solution but (12,4) is not the solution of given equation.

1.5=1/4×1+5/4(which is true)

4=1/4×12+5/4(which is false)

3 0

3 years ago

Read 2 more answers

Please help mee l don’t understand new topic please

jasenka [17]

Answer to first four-

first one- 2f+2f=60

4f=60

f=15

second one-18+2d=38

2d=20

d=10

(short side is 9)

third one-2a+a+5+a+5=26

4a+10=26

a=4

long side is 9

short side is 8

fourth one-3xy+3xy+xy=63

7xy=63

xy=9

long side is 27

4 0

2 years ago

When working problems on the calculator for trig functions, what are you supposed to have your calculator set to?

9966 [12]

Answer:

Degrees

Step-by-step explanation:

dont need one

4 0

3 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 $m_{FG}=\frac{c-0}{a+b-0}$

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

5 0

3 years ago