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2 years ago

Please guys i really Need this please answer this,,

Mathematics

1 answer:

babymother [125]2 years ago

4 0

Answer:

(Roses Ratio) x (The plot of land flowers are planted on Ratio) = (Space the roses occupy), so:

$\frac{1}{5} \times \frac{3}{5} = \frac{3}{25}$

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ABCD is a trapezoid find the value of X and y

Harlamova29_29 [7]

Hey there!

The full area equals . .

Your correct answer would be $x=53,y=106$

Hope this helps.
~Jurgen

6 0

2 years ago

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Write a rule that represents the function (0,0)(1,0.5)(2,2)(3,4.5)(4,8)

kirza4 [7]

The rule that represents those numbers is .....idk

6 0

2 years ago

Maya drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 12 hours. When Maya drove

Roman55 [17]

Answer:

480 miles.

Step-by-step explanation:

Let x represent the distance between Maria's house and mountains ans r represent Maria's rate for going trip.

We have been given that there was heavy traffic on the way there, and the trip took 12 hours.

$\text{Distance}=\text{Rate}\times \text{Time}$

$x=12r$

We are also told that hen Maya drove home, there was no traffic and the trip only took 8 hours. Maria's average rate was 20 miles per hour faster on the trip home.

So Maria's speed while returning back would be $r+20$ .

$x=8(r+20)$

Upon equating both distances, we will get:

$12r=8(r+20)$

$12r=8r+160$

$12r-8r=8r-8r+160$

$4r=160$

$r=\frac{160}{4}$

$r=40$

Upon substituting $r=40$ in equation $x=12r$ , we will get:

$x=12r\\\\x=12(40)\\\\x=480$

Therefore, Maya live 480 miles away from the mountains.

7 0

3 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

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

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

(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

Which ratio is equivalent to 4 : 28?

rusak2 [61]

4 : 28 is the same as 4/28

4/28 = 1/7

The answer is C

3 0

2 years ago

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