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3 years ago

7

Determine the intercepts of the line.

1 answer:

Damm [24]3 years ago

6 0

Y int.: (0,-1.2)
X int.: (-0.7,0)

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In a group of 58 learners at Nakambuda C.S, 34 learners have chosen the field of Agriculture. What fraction of learners will be

asambeis [7]

learners of Agriculture: 34/58

semplify it

divide by 2

34 > 17

58 > 29

they are prime numbers, so can't be semplify anymore

17/29 will do Agriculture

in percentage we have a proportion:

17 : 29 = x : 100

x = (17 * 100) / 29 = 1700 / 29 ≅ 48,62

4 0

3 years ago

What is -2(-y) in expanded form

damaskus [11]

Answer:

2y

Step-by-step explanation:

$(-2)(-y) = (-1)(2)(-1)(y) = (-1)^2\ccdot 2y = 2y$

4 0

2 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

The area is approximately __ square units,.<br><br> use 3.14 for pi

fenix001 [56]

757 units squared. Here is work:

7 0

3 years ago

Debra walked her dog to the south. They walked 50 meters in 150 seconds.

abruzzese [7]

Answer:

B

Step-by-step explanation:

Because... 150 divided by 50 = 3

:)

4 0

3 years ago