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3 years ago

SOMEONE PLEASE HELP ME! I promise i will mark brainlest, but i need the right answer for this. help is much appreciated <3

Mathematics

2 answers:

enot [183]3 years ago

8 0

Answer:

Its D because the x axis can repeat itself

Step-by-step explanation:

A function is a relation in which the members of the domain (x-values) DO NOT repeat. So, for every x-value there is only one y-value that corresponds to it. y-values can be repeated.

e-lub [12.9K]3 years ago

6 0

Answer:

D, the last one

Step-by-step explanation:

A function, f(x), cannot have repeating x-values. Option D has the number 5 repeated twice, {1, 5, 5, 7}, in the x-column, therefore it cannot be a function of x.

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Use a bar model 20-x=6

irinina [24]

Answer:

sdffaf

Step-by-step explanation:

4 0

3 years ago

Question 8

Molodets [167]

D) 50 square units

50 square unitsArea of a square is defined as the number of square units needed to fill a square. In general, the area is defined as the region occupied inside the boundary of a flat object or 2d figure. The measurement is done in square units with the standard unit being square meters (m^2).

the area of square = side^2 square units

and the perimeter of a square = 4 × side units

As per graph, given polygon is a square

WKT,

Side of the square is:

√({5^2+5^2})

= √(50)

= 5√2

Area of the square is:

= a^2

= (5√2)^2

= 25*2

= 50 square units

Correct answer option is:

D) 50 square units

Learn more about squares at : https://brainly.in/question/1761338

#SPJ1

Disclaimer : The correct question is attached below

Question :

The vertices of a polygon are : (0,0), (5,-5), (-5,-5), (0,-10)

Find the area of the polygon with the given coordinates.

A) 10 square units

B) 25 square units

C) 40 square units

D) 50 square units

6 0

2 years ago

Which equation represents the line that passes through the

andreev551 [17]

Answer:

y = 1/5x + 2 1/5

Step-by-step explanation:

first find the negative reciprocal slope of 5x + y = 30 because perpendicular lines have negative reciprocal slope:

=> y = -5x + 30, negative reciprocal of the slope -5 is 1/5

substitute the new slope 1/5 and the coordinate (9, 4) into the equation: y = mx + b to solve for b

=> 4 = 1/5(9) + b

=> 4 = 9/5 + b

=> b = 2 1/5

plug in the slope 1/5 and b = 2 1/5 into the equation: y = mx + b

=> y = 1/5x + 2 1/5

3 0

3 years ago

15.30 find the inverse laplace transform of: 1. (a) f1(s) = 6s 2 8s 3 s(s 2 2s 5) 2. (b) f2(s) = s 2 5s 6 (s 1) 2 (s 4) 3. (c) f

EleoNora [17]

The solution of the inverse Laplace transforms is mathematically given as

$f_{1}(t)=e^{-t}\sin (2 t)$

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$

<h3>What is the inverse Laplace transform?</h3>

Generally, the equation for the function is mathematically given as

$F_{1}(s)=\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}$

By Applying the Partial fractions method

$\frac{6 s^{2}+8 s+3}{s\left(s^{2}+2 s+5\right)}=\frac{A}{s}+\frac{B s+C}{s^{2}+2 s+5}$

$6 s^{2}+8 s+3=A\left(s^{2}+2 s+5\right)+(B s+C) s$

$\begin{aligned}&3=5 A \\&A=\frac{3}{5}\end{aligned}$

Considers s^2 coefficient

$\begin{aligned}&6=A+B \\&B=6 \cdot A \\&B=\frac{27}{5}\end{aligned}$

Consider s coeffici ent

$\begin{aligned}&8=2 A+C \\&C=8-2 A \\&C=\frac{34}{5}\end{aligned}$

Putting these values into the previous equation

$&F_{1}(s)=\frac{3}{5 s}+\frac{27 s+34}{5\left(s^{2}+2 s+5\right)} \\\\&F_{1}(s)=\frac{3}{5 s}+\frac{27(s+1)}{5\left((s+1)^{2}+4\right)}+\frac{7 \times 2}{10\left((s+1)^{2}+4\right)}$

By taking Inverse Laplace Transforms

$f_{1}(t)=\frac{3}{5}+\frac{27}{5} e^{-t} \cos (2t) + \frac{7}{10}\\\\$

$f_{1}(t)=e^{-t}\sin (2 t)$

For B

$F_{2}(s)=\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}$

By Applying Partial fractions method

$\begin{aligned}&\frac{s^{2}+5 s+6}{(s+4)(s+1)^{2}}=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\\\&s^{2}+5 s+6=A(s+1)(s+4)+B(s+4)+C(s+1)^{2}\end{aligned}$

at s=-1

$1-5+6=3 B \\\\B=\frac{2}{3}$

at s=-4

$&16-20+6=9 C \\\\&9 C=2 \\\\&C=\frac{2}{9}$

at s^2 coefficient

1=A+C

A=1-C

A=7/9

inputting Variables into the Previous Equation

$\begin{aligned}&F_{2}(s)=\frac{A}{s+1}+\frac{B}{(s+1)^{2}}+\frac{C}{s+4} \\&F_{2}(s)=\frac{7}{9(s+1)}+\frac{2}{3(s+1)^{2}}+\frac{2}{9(s+4)}\end{aligned}$

By taking Inverse Laplace Transforms

$f_{2}(t)=\frac{7}{9} e^{-t}+\frac{2}{3} e^{-t}+\frac{2}{9} e^{-4 t}$

For C

$F_{3}(s)=\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}$

Using the strategy of Partial Fractions

$\frac{10}{(s+1)\left(s^{2}+4 s+8\right)}=\frac{A}{s+1}+\frac{B s+C}{s^{2}+4 s+8}$

$10=A\left(s^{2}+4 s+8\right)+(B s+C)(s+1)$

S=-1

10=(1-4+8) A

A=10/5

A=2

Consider constants

10=8 A+C

C=10-8 A

C=10-16

C=-6

Considers s^2 coefficient

0=A+B

B=-A

B=-2

inputting Variables into the Previous Equation

$&F_{3}(s)=\frac{2}{s+1}+\frac{-2 s-6}{\left((s+2)^{2}+4\right)} \\\\&F_{3}(s)=\frac{2}{s+1}-\frac{2(s+2)}{\left((s+2)^{2}+4\right)}-\frac{2}{\left((s+2)^{2}+4\right)}$

Inverse Laplace Transforms

$f_{3}(t)=2 e^{-t}-2 e^{-2 t} \cos (2 t)-e^{-2 t} \sin (2 t)$