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3 years ago

Round 1.35 to the nearest whole number.

Mathematics

1 answer:

Yakvenalex [24]3 years ago

5 0

Answer:

Step-by-step explanation:

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Tickets to a school play cost $3 for students and $8 for adults. on opening night, $1840 was collected and 360 tickets sold. use

Pachacha [2.7K]

Answer:

240a (adults) 120s ( students ) hope that helped

Step-by-step explanation:

3/360 8/1840

5 0

3 years ago

Find the total area surface of figure

Neporo4naja [7]

Answer:

$92 \:cm^2$

Step-by-step explanation:

$Surface\:_{area}=2\times(\ell\times w +\ell \times h + w \times h)$

$=2\times(8 \times 2 +8 \times 3 + 2 \times 3)$

$=2 \times (16+24+6)$

$=2\times 46$

$=92\: cm^2$

5 0

2 years ago

Read 2 more answers

What is the slope of the line that passes through the points (-2,7) and (2,-5) in simplest form ?

creativ13 [48]

Its $m=\dfrac{\Delta y}{\Delta x}=\dfrac{7-(-5)}{-2-2}=\dfrac{12}{-4}=\boxed{-3}$ .

Hope this helps.

6 0

3 years ago

Help me with this?<br> ______________

Simora [160]

Answer:

Austin will have to buy 180 squares of carpeting.

Step-by-step explanation:

First find the dimension of the room. We do that by multiplying the width times the length and then subtracting the cut out region in the top right. And, in order to know how big that region we cut out is, we have to do a little subtraction.

We know the room is 18' long on the left side and 12' long on the right side. We subtract 12 from 18 to get 6, and we know that the cut out region is 6' long. We do the same thing with the width, 25' wide at the bottom minus 10' wide at the top and we see that the cut out is 15' wide.

18 x 25 = 450

6 x 15 = 90

450 - 90 = 360. The area of the room is 360 $ft^{2}$

Each piece of carpet is 2' by 1'. So for every 2' long, the piece of carpet is 1' wide. Each carpet piece will cover 2 $ft^{2}$ . Divide 360 by 2 and you get 180.

5 0

3 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below

5 0

3 years ago