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3 years ago

I need the answer and the steps (extra points)

Mathematics

1 answer:

Elena L [17]3 years ago

8 0

To solve the equation, we can first take 17.1 to the right hand side, changing the minus to addition :

1/6x -17.1 = 18

1/6x = -18+17.1

1/6x = -0.9

To take away the fraction, we multiply both sides by 6 so that 1/6 can be taken away to solve the equation:

1/6x = -0.9

1/6x x 6 = -0.9 x 6

6/6x = -3.6

x = -3.6

We can also turn the decimal number into fraction to solve the equation in fraction:

-0.9

= -0.9 x 1/10

= -9/10

To solve the equation:

1/6x = -9/10

x = -9/10 x 6

x = -36/10

x = -18/5

x= -3 3/5

Therefore, x = -3.6 or -3 3/5.

Hope it helps!

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Can I please have help

Dima020 [189]

Answer:

15a+12ac+6ab

Step-by-step explanation:

3a(5 + 4c + 2b)

5*3a = 15a

4c*3a = 12ac

2b*3a = 6ab

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2 years ago

Read 2 more answers

Mr Lim read 3 books in 2 hours. He took the same amount of time to read the first two books. He took 15 minutes longer to read t

Naya [18.7K]

He took 75 minutes to read the third book.

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3 years ago

Use properties to find the sum or product 8x51

SCORPION-xisa [38]

$8\times51=8\times(50+1)=8\times50+8\times1=400+8=408$

3 0

3 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

3 years ago

Carrie is driving to a friend’s house for the weekend. The friend lives in a town 160 mi. away. So far, Carrie has driven 64 mi.

ankoles [38]

Divide the amount she drove by the total amount she needs to drive, them multiply that answer by 100 to make it a percent. This will give you the percentage she drove. To find the percentage she has left,, subtract from 100%.

64 / 160 = 0.4

0.4 * 100 = 40% ( percent she has already driven)

100% - 40% = 60% left.

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3 years ago