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2 years ago

8. Is line / parallel to line m? Explain. (1 point)

Mathematics

1 answer:

puteri [66]2 years ago

7 0

Answer:

what lines..? post picture.

Step-by-step explanation:

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Which of the following is a radical equation?<br> Ox√3=13<br> O x+√3=13<br> O√x+3=13<br> O x+3=√13

sashaice [31]

Answer:

The answer would be: √x+3=13

Step-by-step explanation:

A radical equation is an equation with at least one variable put on a radical form. In this question, there is only one kind of variable which is x. Then, you need to find the equation where x is in the radical form. The radical form can be expressed with square root symbol.

The option with √x would be only √x+3=13

8 0

1 year ago

What is the coefficient in the expression 40a-61=17

zepelin [54]

Answer:

a = 1.95

Step-by-step explanation:

40a - 61 = 17

40a - 61 + 61 = 17 + 61

40a = 78

40a / 40 = 78 / 40

a = 1.95

5 0

3 years ago

William's typical running speed is 6 mi/h. The difference between his fastest speed and his typical speed is 1.5 min. The equati

Over [174]

Answer:

D) 7.5 mi/h. hcbhcvu yc yv. y yv yh

6 0

2 years ago

Read 2 more answers

A shop advertised a 27% off sale. If a coat originally sold for 253 , find the decrease in price and the sale price

OverLord2011 [107]

253 * 0.27= 68.31 (discounted price)

253-68.31 OR 253*0.73= 184.69 (sale price)

8 0

3 years ago

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

GREYUIT [131]

Answer:

$y=\dfrac{1}{1-Ke^{-t}}$

Step-by-step explanation:

Given

The given equation is a differential equation

$\dfrac{dy}{dt}=y-y^2$

$\dfrac{dy}{dt}=-(y^2-y)$

By separating variable

⇒ $\dfrac{dy}{(y^2-y)}=-t$

$\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt$

Now by taking integration both side

$\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt$

⇒ $\ln (y-1)-\ln y=-t+C$

Where C is the constant

$\ln \dfrac{y-1}{y}=-t+C$

$\dfrac{y-1}{y}=e^{-t+c}$

$\dfrac{y-1}{y}=Ke^{-t}$

$y=\dfrac{1}{1-Ke^{-t}}$

from above equation we can say that

When t will increases in positive direction then $e^{-t}$ will decreases it means that ${1-Ke^{-t}}$ will increases, so y will decreases. Similarly in the case of negative t.

4 0

3 years ago