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3 years ago

Find the value of the variable plz help

Mathematics

2 answers:

Naddika [18.5K]3 years ago

7 0

Answer:

Step-by-step explanation:

Vertically opposite angles are equal.

8y + 36 = 14y -24 Subtract 36 to both sides

8y = 14y - 24 - 36 Combine

8y = 14y - 60 Subtract 14y from both sides

8y - 14y = - 60

-6y = -60 Divide by - 6

-6y/-6 = -60/-6

y = 10

===========================

x +48 = 64 Subtract 48 from both sides

x +48 - 48 = 64-16

x = 16

kari74 [83]3 years ago

6 0

Answer:

x = 16

y = 10

Step-by-step explanation:

*first of, we know that one angle is 64 degrees, which means the opposite side is also 64 degrees

we can find x by the following formula:

⇒ 64 = 48 + x

then subtract 48 on both sides:

⇒ 64 - 48 = 48 - 48 + x

⇒ 16 = x

we now know that two of the angles are 64 degrees, this means that the other two angles add up to 116 degrees (180 - 64 = 116)

with this being said, lets talk a look at (8y + 36)

we can solve y from the following equation:

⇒ 116 = 8y + 36

first, subtract 36 on both sides

⇒ 116 - 36 = 8y + 36 - 36

⇒ 80 = 8y

then divide 8 on both sides

⇒ $\frac{80}{8} = \frac{8y}{8}$

⇒ 10 = y

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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:

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3 years ago

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defon

Answer:

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Step-by-step explanation:

Ratio is 30 : 4200

Divide both numbers by 30:-

= 1 : 140 (answer)

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