Plz hurry and answer i will give brainliest<br><br> what is 2*2+6/2*3

A rectangular floor is 6 yards long and 4 yards wide. What is the area of the floor in square feet? Please help!

lys-0071 [83]

Area = length x width

A = L•W

There are 3 feet in one yard.

6 yards = 18 feet

4 yards = 12 feet

A = (18)(12)

A = 216 square feet

8 0

3 years ago

Three points, A, B, and C exists in space such that B is "between" A and C. It is known that AB=7, BC=4, and AC=9. Are points A,

sergeinik [125]

If the points are collinear, then that means they lie on the same line when connected to each other. In order to prove this, the distances between the points must sum up to the length of the line.

AB + BC = AC
7 + 4 ? 9
11 ≠ 9

Since they do not sum up, therefore, the points are not collinear.

7 0

3 years ago

Read 2 more answers

What's the answer to this? And how do u do it

andrew-mc [135]

-5n - 3 - 3 > 19
-5n - 6 > 19
+ 6 +6
-5n > 25
divide by -5 on each side
change signs since dividing by negative

N< -5

For graphing on the line, you will create an OPEN circle on -5 and sketch it to the left.

3 0

3 years ago

Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y-axis the region under the curve y = 1

velikii [3]

Using the shell method, the volume is given exactly by the definite integral,

$2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx$

Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the $i$ -th subinterval

$\ell_i=\dfrac{i-1}5$

$r_i=\dfrac i5$

where $1\le i\le5$ . The midpoint of each subinterval is

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}$

Then the Riemann sum approximating the integral above is

$2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5$

$\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)$

$\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}$

(compare to the actual value of the integral of about 14.45)

3 0

3 years ago

Jason and Britton are driving to St. George. Jason got a 20 mile head start and drove an

harina [27]

Answer:

2 hours, 150 miles

Step-by-step explanation:

The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.

<h3>Separation distance</h3>

Jason got a head start of 20 miles, so that is the initial separation between the two drivers.

<h3>Closure speed</h3>

Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.

<h3>Closure time</h3>

The relevant relation is ...

time = distance/speed

Then the time it takes to reduce the separation distance to zero is ...

closure time = separation distance / closure speed = 20 mi / (10 mi/h)

closure time = 2 h

Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.

__

<em>Additional comment</em>

The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.

3 0

1 year ago

Plz hurry and answer i will give brainliest what is 2*2+6/2*3

Plz hurry and answer i will give brainliest what is 22+6/23