What is the value of the expression below when z=4 3z2-10z-5

Barba can walk 3200 meters in 24 minutes.How far can she walk on 3 mins

SashulF [63]

Answer:

400 meters in 3 minutes

Step-by-step explanation:

First find how far she can walk in 1 minute.

3200/24 = 133 1/3.

She can walk 133 1/3 meters in 1 minute. If we want to find out how far she can walk in 3 minutes. Multiply that number by 3.

133 1/3 * 3 = 400.

Hope that helps.

8 0

2 years ago

When ∆A′B′C′ is reflected across the line x = -2 to form ∆A″B″C″, vertex (note: please enter your response using apostrophes ins

maria [59]

(A) because it is reflecting over the x-axis.

7 0

3 years ago

Will give brainliest Please help with problem, I don't understand it

Alex787 [66]

Answer:

111.375

Step-by-step explanation:

each triangle are equal to each other, it equals (b x h)/2

good luck and thanks for the brainliest!

6 0

2 years ago

The first inverse operation you would use to solve the equation 2+5x= -8 for x is subtraction. true or false?

jok3333 [9.3K]

Answer:

It would be True

Step-by-step explanation:

2+5x = -8

SUBTACT 2 on both sides and it would look like this:

5x = -8 - (2)

5X= -10

then DIVIDE 5 both side to get "X" ALONE,

x= -10/5

then divide -10 with 5

equal to:

x = -2

4 0

2 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago