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3 years ago

PLEASE HELP!!! WILL MARK BRANLIEST

Mathematics

2 answers:

BartSMP [9]3 years ago

8 0

Answer:

I'm glad you asked!

Step-by-step explanation:

You multiply all the numbers.

$10x4x5=2000$

Your answer is 2000 <em>ft.</em>

Alexxx [7]3 years ago

7 0

Answer:

83.33

Step-by-step explanation:

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Please help!

lana [24]

Given Information:

Principle amount = P = $6,000

Interest rate = r = 4% = 0.04

Period in years = t = 5

Required Information:

How much interest will he earn in 5 years = ?

Answer:

Amount of interest = $1,299.92

Step-by-step explanation:

Using the formula given in the question,

$B = P(1 + r )^{t}$

Where B is the final amount, P is the initial amount, r is the interest rate and t is the number of years

$B = P(1 + r )^{t}\\\\B = 6,000(1 + 0.04)^{5}\\\\B = 7,299.92$

The amount of interest earned is

$Amount \:of\: interest = B - P \\\\Amount \:of \:interest = 7,299.92 - 6,000\\\\Amount \:of \:interest = 1,299.92$

Therefore, Quincy has earned $1,299.92 in terms of interest by investing $6,000 in a savings account at the rate of 4% annual interest for a period of 5 years.

6 0

3 years ago

Which of the choices below follow an exponential pattern? Select all that apply.

Masja [62]

What are the choices?

8 0

4 years ago

Find the five arithmetic means between -18 and 36

Andreas93 [3]

Yeah it really be like that sometimes, you gotta try girl

5 0

3 years ago

Find an equation of the plane with x-intercept a, y-intercept b, and z-intercept c. What is the distance between the origin and

Elenna [48]

Answer:

The equation is:

(1/a)x + (1/b)y + (1/c)z = 1

Step-by-step explanation:

The direction vector between the points (a, 0, 0) and (0, b, 0) is given as:

<0 - a, b - 0, 0 - 0>

<-a, b, 0> .....................(1)

The direction vector between (0, 0, c) and (0, b, 0) is given as:

<0 - 0, b - 0, 0 - c>

= <0, b, -c> .....................(2)

To obtain the direction vector that is normal to the surface of the plane, we take the cross product of (1) and (2).

Doing this, we have:

<-a, b, 0> × <0, b, -c> = <-bc, -ac, -ab>

To find the scalar equation of the plane we can use any of the points that we know. Using (0, b, 0), we have:

(-bc)x + (-ac)y + (-ab)z = (-bc)0 + (-ac)b + (-ab)0

(bc)x + (ac)y + (ab)z = (ac)b

Dividing both sides by abc, we have:

(1/a)x + (1/b)y + (1/c)z = 1

3 0

3 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago