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3 years ago

89 more than the quotient of an unknown number and 62 is 94

1 answer:

6 0

X/62 + 89 = 94

x/62 +89 = 94
 - 89 - 89
x/62 = 5
* 62 *62
x = 310

To check:
quotient of an unknown number and 62
310/62 = 5

5 + 89 = 94
94 = 94

5 is the quotient of an unknown number and 62.

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Doss [256]

Answer:

2/3

Step-by-step explanation:

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3 years ago

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3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

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2 years ago

Which table shows a constant rate of change?

Dafna1 [17]

Answer:

B. 2nd Table

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Which two points should the line of best fit go through to best represent the data in the scatterplot?

Ilya [14]

Given : A graph with points (1, 3), (2, 7), (3, 8), (4, 11), (5, 12), (6, 13).

To find : From Which two points the line of best fit goes through

Solution:

(3, 8) & (5, 12) will be two points

as then line points will be

( 1 , 4 ) , ( 2, 6) , ( 3 , 8 ) , ( 4, 10) , (5 . 12) & ( 6 , 14)

(1, 3) will be below ( 1 , 4 )

(2 , 7) will be above ( 2 , 6)

( 3 , 8 ) will be on line

(4 , 11 ) will be above ( 4 , 10)

( 5 , 12 ) will be on line

( 6 , 13) will be below ( 6 , 14 )

(1, 3) & ( 6 , 13) two points are below line

(2 , 7) & (4 , 11 ) two points above line

Hence points are Equally scattered above & below line

Hence (3, 8) & (5, 12) will be two points on line of best fit

Hope it helpss !!

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3 years ago

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