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3 years ago

Helpppppp ??????????

Mathematics

1 answer:

Rzqust [24]3 years ago

5 0

3/4 and 1/4 is the simplest way you can put it with 3/4 being when he walks.

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4.2z + 3.8z simplify the expression (Use the operation symbols in the math palette as needed. Simplify your answer. Use whole n

nata0808 [166]

A picture pls I need it

3 0

3 years ago

Find dy/dx of y=csc(square root of x)

Vitek1552 [10]

Answer:

$y' = -\dfrac{\cot x \csc x}{2 \sqrt{x}}$

Step-by-step explanation:

y = csc x

y' = -cot x csc x

$y = \csc \sqrt{x}$

$y' = \dfrac{d}{dx} [\csc \sqrt{x}]$

$y' = (-\cot x \csc x) \dfrac{d}{dx} \sqrt{x}$

$y' = (-\cot x \csc x) \dfrac{d}{dx} x^{\frac{1}{2}}$

$y' = (-\cot x \csc x) \dfrac{1}{2} x^{-\frac{1}{2}}$

$y' = -\dfrac{\cot x \csc x}{2 \sqrt{x}}$

5 0

3 years ago

Evaluate the expression when c=-5.1 and d= - 7.8.

FrozenT [24]

Answer:

−31.5

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

If the line joining the points (2,y) and (4,6) has the slope -3/2 find the value of y

Yanka [14]

Answer: y=9

Step-by-step explanation:

slope= $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$

$\frac{-3}{2} =\frac{6-y}{4-2}$

$-6=2(6-y)$

$-6=12-2y$

$2y=18$

$y=9$

4 0

4 years ago

If the maximum acceleration that is tolerable for passengers in a subway train is 1.74 m/s2 and subway stations are located 930

Free_Kalibri [48]

Given that the subway stations are 930 m apart, the train have to be accerelated for half the distance and then decerelated for the rest of the distance.

Recall that the distance travelled by an object with an initial velocity, u, for a period of time, t, at an accereration, a, is given by
$s=ut+ \frac{1}{2} at^2$

But, we assume that the train accelerates from rest, thus
$s=\frac{1}{2} at^2 \\ \\ \Rightarrow465=\frac{1}{2}(1.74)t^2 \\ \\ \Rightarrow t^2=534.48 \\ \\ \Rightarrow t=\sqrt{534.48}=23.12$

The maximum speed is attained at half the center of the distance between subway stations (i.e. at distance = 465 m).

Thus, maximum speed = distance / time = 465 / 23.12 = 20.11 m/s.

6 0

4 years ago