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3 years ago

Can some one please help me i don't understand this.

Mathematics

1 answer:

omeli [17]3 years ago

3 0

People with the links be annoying these days bro who even Invented them

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Factorise x 2 +3x - 4

AysviL [449]

Answer:

5x - 4 would be your answer.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Pls help fast will get brainliest

tatuchka [14]

Answer: (3.75,5.5)
Hope this helps

4 0

3 years ago

a movie theater charges $9.50 per ticket for adults and $4.50 per ticket for children. Write an expression that represents the t

Alborosie

X and y would equal the same so 23

7 0

3 years ago

A pool estimate that 43% of likely voters are in favor of additional restrictions on teen drivers the Poll has a margin of error

MissTica

Answer:

Kindly check explanation

Step-by-step explanation:

Given that :

Estimated number of voters in favor of additional restriction = 43%

Margin of Error = ±4%

Write and solve the absolute value equations to find the minimum and maximum percent of voters that support the restriction:

Minimum percentage of voters = 4% lesser than the estimated number of voters

Maximum percentage of voters = 4% greater than the estimated number of voters

Let number of voters = v

|v - 43%| = ±4%

v = - 4% + 43% = 39% = minimum percentage

v = +4% + 43% = 47% = maximum percentage

39% ≤ v ≤ 47%

6 0

4 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

4 years ago