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3 years ago

How do you turn 0.875 into a fraction

Mathematics

1 answer:

galina1969 [7]3 years ago

6 0

Answer: 5/8

Step-by-step explanation:

0.875 is equal to 875/1000 because there are 3 decimal places, which is the amount of zeroes after the 1 as i like to think of it, and then to simplify it

875/1000 = 35/40 = 5/8

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Soulotion to equation if y is =15 when x=2 ,find y when x=8

skelet666 [1.2K]

Answer:

The value of y for x = 8 is 60 .

Step-by-step explanation:

Given expression as :

For x = 2 , y = 15

Let the equation be written with constant k'

So, y = k x

Or, k = $\dfrac{y}{x}$

∴ k = $\dfrac{15}{2}$

<u>Now, again </u>

For x = 8

equation can be written as

y = k x

Or, y = $\dfrac{15}{2}$ × 8

Or, y = $\frac{15\times 8}{2}$

∴ y = 15 × 4

I.e y = 60

Hence The value of y for x = 8 is 60 . Answer

5 0

3 years ago

Enzo is studying the black bear population at a large national park.

Naddik [55]

Answer:

2,973

Step-by-step explanation:

The black bear population B(t), in the park is modeled by the following function:

$B(t) = 2500 \cdot 2^{0.01t}$

Where t is the time(in years) elapsed since the beginning of the study.

We want to determine the black bear population in 25 years time, t=25.

$B(25) = 2500 \cdot 2^{0.01*25}\\=2973.02\\\approx 2973$

There will be 2,973 black bears in 25 years time.

4 0

3 years ago

Determine if AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯ are parallel, perpendicular, or neither. Given: A (−1, 3), B (0, 5), C (2, 1), D (6, −1)

dolphi86 [110]

Answer:

perpendicular

Step-by-step explanation:

To determine if AB and CD are parallel, perpendicular, or neither, we need to get the slope of AB and CD first

Given A (−1, 3), B (0, 5),

Slope Mab = 5-3/0-(-1)

Mab = 2/1

Mab = 2

Slope of AB is 2

Given C (2, 1), D (6, −1)

Slope Mcd = -1-1/6-2

Mcd = -2/4

Mcd = -1/2

Slope of CD is -1/2

Take their product

Mab * Mcd = 2 * -1/2

Mab * Mcd = -1

Since the product of their slope is -1, hence AB and CD are perpendicular

5 0

2 years ago

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.