1o) Em um posto de combustível o litro da gasolina custa R$ 4,68. Paulo colocou 8 litros de gasolina. Quanto ele pagou?
a) R$ 41,63
b) R$ 38,52
c) R$ 37,44
d) R$ 32,36
e) R$ 25,10
2o) Flávia foi à papelaria e comprou uma régua por R$ 3,80, uma borracha por R$ 1,35 e duas canetas por
R$ 1,90 cada uma.
Se Flávia deu uma nota de R$ 20,00 para pagar a compra, quanto recebeu de troco?
a) R$ 12,95
b) R$ 12,10
c) R$ 11,60
d) R$ 11,05
e) R$ 10,90
3o) Cinco atletas participaram da final de uma corrida de 100 metros na Escola Voo Rasteiro.
A tabela abaixo apresenta o tempo que cada um levou para terminar a prova.
Atleta Tempo (segundo)
Daniel 12,5
Fernando 10,3
Manoel 10,5
Arnaldo 11,4
Júnior 13,9
Quem chegou em primeiro lugar nessa corrida?
a) Daniel
b) Fernando
c) Manoel
d) Arnaldo
e) Júnior
4o) (Obmep) Alvimar pagou uma compra de R$ 3,50 com uma nota de R$ 5,00 e recebeu o troco em
moedas de R$ 0,25. Quantas moedas ele recebeu?
a) 4
b) 5
c) 6
d) 7
e) 8

for any of you who need the translation here it is

step-by-step explanation:

find the angular size of the equatorial radius of the earth, if the horizontal parallax of the sun

square root

step-by-step explanation:

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.