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2 years ago

Solve the following expression when v = 10 and e= 5 2v ÷ e + 3

Mathematics

1 answer:

Montano1993 [528]2 years ago

6 0

Hello!

Answer:

5/2 or 2.5

Step-by-step explanation:

<em>Plug in the numbers for v and e </em>

2(10)÷5+3

20÷8

5/2 or 2.5

Hope this helps!

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Beverly is starting a new diet. Her current weight is 160 pounds. She expects to lose 4 pounds per month. If x represents the nu

nevsk [136]

The answer is B) or y = -4x + 160

If her initial weight is 160 pounds, then you must subtract 4x, representing how much weight she lost over time, to find her current weight y

But, since y = 160 – 4x isn't an option, we can reverse to negatives as long as you change the addition/subtraction sign

So, subtracting positive 4x is the same as adding -4x

Therefore, y = 160 – 4x is the same as y = -4x + 160

4 0

3 years ago

Read 2 more answers

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

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1 year ago

The manager at the local auto shop has found that the probability that a car brought into the shop requires an oil change is 0.8

Ber [7]

Answer:

0.85

Step-by-step explanation:

Given two events A and B, the probability that either A or B occurs is given by:

$p(A\cup B) = p(A)+p(B)-p(A\cap B)$

where

$p(A)$ the probability that A occurs

$p(B)$ is the probability that B occurs

$p(A\cap B)$ is the probability that both A and B occur at the same time

In this problem, we know the following facts:

$p(o) = 0.83$ is the probability that the car requires an oil change

$p(b)=0.17$ is the probability that the car requires a brake repair

$p(o\cap b) = 0.15$ is the probability that the car requires both an oil change and brake repair

Therefore, the probability that either o (car requiring oil change) or b (car requiring brake repait) occur is:

$p(o\cup b)=p(o)+p(b)-p(o\cap b)=0.83+0.17-0.15 = 0.85$

4 0

3 years ago

Plssss I need help Im in classsss plsss

Sav [38]

Step-by-step explanation:

Y=-4x+3

2x+3y=19

2x+3(-4x+3)=19

2x-12x+9=19

-10x=10

X=-1

y=-4*(-1)+3=4+3=7

x=-1 y=7

5 0

3 years ago

Whats 60 percent of 80

Degger [83]

48 all you had to do was out it in a calculator that's what I did

8 0

3 years ago

Read 2 more answers

Solve the following expression when v = 10 and e= 5 2v ÷ e + 3​

Solve the following expression when v = 10 and e= 5 2v ÷ e + 3