All categories

3 years ago

Area of a 90in 3yd rectangle

1 answer:

5 0

Hello,

Converting 90 in to yd, we have:
90in * 1ft/12in * 1yd/3ft = 2.5 yd

Next,
Since A of a rectangle is b*h,
A = 2.5yd * 3yd = 7.5 yd ^2.

Good luck to you!

You might be interested in

In a bag there are 3 red marbles, 2 yellow marbles, and 1 blue marble. what is the probability in simplest form of a yellow marb

dangina [55]

Answer:

The probability is 1/3, or in percent form, 33.3%.

Step-by-step explanation:

First, add up all the marbles. 3 plus 2 plus 1 equals 6 total marbles.

There are 2 yellow marbles out of 6 total marbles. Make it a fraction: 2/6.

Simplify the fraction by dividing by 2, to get 1/3, which is 33.3%.

Hope this helps! Feel free to give me Brainliest if you feel this helped. Have a good day, and good luck on your assignment. :)

5 0

3 years ago

What is the remandier to 92931 divided by 13?

ivann1987 [24]

No, 92932 is not divisible by 13 because it has to much numbers for getting the answer

5 0

4 years ago

Write a function rule for the following geometric sequence.

Natasha2012 [34]

I think it f(x) = 3x + 4 if I am wrong I am so so sorry have a great day.

3 0

3 years ago

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p

GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

$t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2$

you replace t1 from the first equation in the second equation:

$(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}$

Then, for t2 = 0 and t2=1/2 you obtain for t1:

$t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2$

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths intersect are:

$r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)$

B) You have the following two trajectories of two independent particles:

$r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)$

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

$t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t$

From the first equation you have:

$t=1+2t\\\\t=-1$

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0

3 years ago

After 6 hours, a driver had traveled 390 miles. how far had the driver traveled after 4 hours

lozanna [386]

Sorry I needed points for an important quiz

8 0

3 years ago