Answer questions 1, 2, 3 and challenge

Mathematics

1 answer:

hjlf4 years ago

4 0

1. find volume
V=LWH
V=25*13*14
V=4550
each covers 1000
how many will cover the whole thing?
we can't put 4 becuase that leaves 550 uncovered
add another

5 is answer (it is too much , but at leas not too little) answer is D

2.
V=LWH
V=6*4*10
V=240

answer is D

3. count how many are there, don't forget the hidden ones
13 cubes
volume=legnth tiems width times height
cubes so
v=side^3
side=3
3^3=27
13 cubes
13*27=351 in^3

challenge (this is meant to challenge you, not for you to ask people, but I will solve anyway, robbin you of experience)

V=LWH
V=5880
H=30
5880=LW30
divide both sides by 30
196=LW
they have square bases so L=W
196=L^2
sqrt both sides
14=L=W
the side legnth is 14in

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

almond37 [142]

Answer: 742

<u>Step-by-step explanation:</u>

The given sequence {27, 31, 35, ... } provides the following information:

the first term (a₁) = 27
the difference (d) = 4

We can use the information above to find the explicit rule of the sequence:

$a_n=a_1+d(n-1)\\.\quad =27+4(n-1)\\.\quad =27+4n-4\\.\quad =23+4n$

We can use the explicit rule to find the 14th term (a₁₄)

$a_n=23+4n\\a_{14}=23+4(14)\\.\quad =23+56\\.\quad =79$

Next, we can input the first and last term of the sequence into the Sum formula:

$S_n=\dfrac{a_1+a_n}{2}\times n\\\\S_{14}=\dfrac{a_1+a_{14}}{2}\times 14\\\\.\quad =\dfrac{27+79}{2}\times 14\\\\.\quad =\dfrac{106}{2}\times 14\\\\.\quad =53\times 14\\\\.\quad =742$