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3 years ago

1. A student has several boxes of penils in his backpack. Each box contains 12 pencils. The student

Mathematics

2 answers:

arsen [322]3 years ago

6 0

The answer is a. y=12x+3

romanna [79]3 years ago

5 0

Answer:

I think it is A because 12 is the slope and 3 is just b... try y = mx + b

Step-by-step explanation:

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See the picture below and answer for 50 points and brainliest

Luden [163]

Answer:

Non-linear equations:

$y=x^2 +2x-3$

$3y=2^x$

Step-by-step explanation:

Linear equation: y = mx + b

Therefore, a linear equation has 2 variables that have no exponents.

So the non-linear equations are:

$y=x^2 +2x-3$ (this is a quadratic equation - polynomial with degree 2)

$3y=2^x$ (this is an exponential equation)

8 0

2 years ago

For questions 10-11 if l || m, find the values of x and y.

klemol [59]

Answer:

10) 9x - 2° = 5x + 54° (corresponding angles are equal)

9x - 5x = 54 + 2

4x = 56

x = 56/4

x = 14°

10y + 6° = 9x - 2° (linear pair)

10y + 6° = 9(14)° - 2°

10y + 6° = 126° - 2°

10y + 6° = 124°

10y = 124° - 6°

10y = 118

y = 118/10

Sorry, i don't know how to do the 11th question

but hope this helps you!

8 0

2 years ago

The dot plot below shows how many apple slices each girl at kaylas party ate which of the following represents the mean of the a

katovenus [111]

The answer to this question is 5.

8 0

3 years ago

In the rectangle below ac=48 units what is de

r-ruslan [8.4K]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

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