Respuesta :
1 is 8w.
2 is X = 18.
3 is X = 7.
4 is X = 3.
5 is 0.25q + 0.05n.
6 is D.
7 is B.
8 is 17.29.
2 is X = 18.
3 is X = 7.
4 is X = 3.
5 is 0.25q + 0.05n.
6 is D.
7 is B.
8 is 17.29.
ANSWER TO QUESTION 1
We were given that, Carlo recycles 8 glass bottles every week.
That means in 2 weeks, Carlo will recycle,
[tex]8 \times 2 = 16 \: glass \: bottles[/tex]
in 3 weeks time, Carlo will recycle,
[tex]8 \times 3 = 24 \: glass \: bottles[/tex]
Therefore w weeks time, Carlo will recycle,
[tex]8 \times w = 8w \: \: \: glass \: \: bottles.[/tex]
The correct answer is C.
ANSWER TO QUESTION 3.
We want to solve the linear equation,
[tex]x + 9.5 = 27.5[/tex]
This means we need to isolate x, so that it will be on one side of the equation.
To do that, we must first add the additive inverse of
[tex]9.5[/tex]
to both sides of the equation.
This implies that,
[tex]x + 9.5 + - 9.5 = 27.5 + - 9.5[/tex]
This gives us,
[tex]x + 9.5 - 9.5 = 27.5 - 9.5[/tex]
This simplifies to,
[tex]x + 0= 18[/tex]
Therefore,
[tex]x = 18[/tex]
The correct answer is B.
ANSWER TO QUESTION 4.
The given linear equation is
[tex]9x = 63[/tex]
We need to make x the subject, by multiplying both sides of the equation by multiplicative inverse or the reciprocal of 9.
This implies that,
[tex] \frac{1}{9} \times 9x = \frac{1}{9} \times 63[/tex]
We simplify to obtain,
[tex]x = 7[/tex]
Therefore you enter the whole number 7, in the blank space provided.
ANSWER TO QUESTION 5.
We want to find the numerical solution to
[tex] \frac{1}{21} x = \frac{1}{7} [/tex]
Clear the fraction by multiplying through by the least common multiple of 7 and 21 which is 21.
This implies that,
[tex] 21 \times \frac{1}{21} x = \frac{1}{7} \times 21[/tex]
This simplifies to give us,
[tex]x = 3[/tex]
The correct answer is A.
ANSWER TO QUESTION 6
We want to convert
[tex]q[/tex]
quarters and
[tex]n[/tex]
nickels to dollars.
Recall that,
[tex]1 \: dollar = 4 \: \: quarters[/tex]
[tex]\Rightarrow \: \frac{1}{4} dollars = 1 \: quarters[/tex]
[tex]\Rightarrow \: 0.25q \: \: dollars = q \: quarters[/tex]
Also,
[tex]1 \: \: dollar = 20 \: \: nickels[/tex]
[tex] \frac{1}{20} \: \: dollar = 1\: \: nickel[/tex]
[tex] 0.05n\: \: dollars = n\: \: nickels[/tex]
This implies that,
[tex]q \: \: quarters \: and \: n \: \: nickels =( 0.25q + 0.05n) \: dollars[/tex]
The correct answer is C.
ANSWER TO QUESTION 7.
We were told that the students must collect at least $120.
If
[tex]x[/tex]
represents the dollars the students must collect for the fair, then write the inequality,
[tex]x \geqslant 120[/tex]
If the students have already collected $30.
Then the amount remaining for the students to collect is at least $90.
The correct answer is B.
See graph in attachment.
ANSWER TO QUESTION 8
We were told that the store sold 60 copies of a magazine for $240. This means that,
[tex]each \: copy = \frac{240}{60} [/tex]
[tex]each \: copy = 4 \: \: dollars[/tex]
This implies that,
[tex]x[/tex]
copies will cost,
[tex]4x \: \: dollars[/tex]
Therefore the equation that best represents the price, in dollars,
is
[tex]y = 4x[/tex]
and the ordered pairs
[tex](1,4),(2,8),(3,12)[/tex]
The correct answer is B.
ANSWER TO QUESTION 9
The total amount of money Aura has is $71.
The total money spent
[tex] = 27.58 + 26.13 \: \: dollars[/tex]
[tex] = 53.71 \: \: dollars[/tex]
The amount remaining
[tex] = 71 - 53.71 = 17.29 \: dollars[/tex]
Therefore Aura spent $17.29 in buying a pair of shoes.
ANSWER TO QUESTION 10
The inequality that models Minni's situation is
[tex]0.35x + 0.99y + 0.59 \leqslant 4[/tex]
Whether Minni can buy 2 stickers and 2 erasers or not, we need to substitute
[tex]x = 2 \: \: and \: \: y = 2[/tex]
in to the inequality.
If the two values satisfy the inequality, then it is possible.
[tex]0.35(2) + 0.99(2) + 0.59 \leqslant 4[/tex]
This gives us,
.
[tex]0.7 + 1.98+ 0.59 \leqslant 4[/tex]
[tex]3.27 \leqslant 4[/tex]
The above statement is true, therefore it is possible.
The correct answer is A.
We were given that, Carlo recycles 8 glass bottles every week.
That means in 2 weeks, Carlo will recycle,
[tex]8 \times 2 = 16 \: glass \: bottles[/tex]
in 3 weeks time, Carlo will recycle,
[tex]8 \times 3 = 24 \: glass \: bottles[/tex]
Therefore w weeks time, Carlo will recycle,
[tex]8 \times w = 8w \: \: \: glass \: \: bottles.[/tex]
The correct answer is C.
ANSWER TO QUESTION 3.
We want to solve the linear equation,
[tex]x + 9.5 = 27.5[/tex]
This means we need to isolate x, so that it will be on one side of the equation.
To do that, we must first add the additive inverse of
[tex]9.5[/tex]
to both sides of the equation.
This implies that,
[tex]x + 9.5 + - 9.5 = 27.5 + - 9.5[/tex]
This gives us,
[tex]x + 9.5 - 9.5 = 27.5 - 9.5[/tex]
This simplifies to,
[tex]x + 0= 18[/tex]
Therefore,
[tex]x = 18[/tex]
The correct answer is B.
ANSWER TO QUESTION 4.
The given linear equation is
[tex]9x = 63[/tex]
We need to make x the subject, by multiplying both sides of the equation by multiplicative inverse or the reciprocal of 9.
This implies that,
[tex] \frac{1}{9} \times 9x = \frac{1}{9} \times 63[/tex]
We simplify to obtain,
[tex]x = 7[/tex]
Therefore you enter the whole number 7, in the blank space provided.
ANSWER TO QUESTION 5.
We want to find the numerical solution to
[tex] \frac{1}{21} x = \frac{1}{7} [/tex]
Clear the fraction by multiplying through by the least common multiple of 7 and 21 which is 21.
This implies that,
[tex] 21 \times \frac{1}{21} x = \frac{1}{7} \times 21[/tex]
This simplifies to give us,
[tex]x = 3[/tex]
The correct answer is A.
ANSWER TO QUESTION 6
We want to convert
[tex]q[/tex]
quarters and
[tex]n[/tex]
nickels to dollars.
Recall that,
[tex]1 \: dollar = 4 \: \: quarters[/tex]
[tex]\Rightarrow \: \frac{1}{4} dollars = 1 \: quarters[/tex]
[tex]\Rightarrow \: 0.25q \: \: dollars = q \: quarters[/tex]
Also,
[tex]1 \: \: dollar = 20 \: \: nickels[/tex]
[tex] \frac{1}{20} \: \: dollar = 1\: \: nickel[/tex]
[tex] 0.05n\: \: dollars = n\: \: nickels[/tex]
This implies that,
[tex]q \: \: quarters \: and \: n \: \: nickels =( 0.25q + 0.05n) \: dollars[/tex]
The correct answer is C.
ANSWER TO QUESTION 7.
We were told that the students must collect at least $120.
If
[tex]x[/tex]
represents the dollars the students must collect for the fair, then write the inequality,
[tex]x \geqslant 120[/tex]
If the students have already collected $30.
Then the amount remaining for the students to collect is at least $90.
The correct answer is B.
See graph in attachment.
ANSWER TO QUESTION 8
We were told that the store sold 60 copies of a magazine for $240. This means that,
[tex]each \: copy = \frac{240}{60} [/tex]
[tex]each \: copy = 4 \: \: dollars[/tex]
This implies that,
[tex]x[/tex]
copies will cost,
[tex]4x \: \: dollars[/tex]
Therefore the equation that best represents the price, in dollars,
is
[tex]y = 4x[/tex]
and the ordered pairs
[tex](1,4),(2,8),(3,12)[/tex]
The correct answer is B.
ANSWER TO QUESTION 9
The total amount of money Aura has is $71.
The total money spent
[tex] = 27.58 + 26.13 \: \: dollars[/tex]
[tex] = 53.71 \: \: dollars[/tex]
The amount remaining
[tex] = 71 - 53.71 = 17.29 \: dollars[/tex]
Therefore Aura spent $17.29 in buying a pair of shoes.
ANSWER TO QUESTION 10
The inequality that models Minni's situation is
[tex]0.35x + 0.99y + 0.59 \leqslant 4[/tex]
Whether Minni can buy 2 stickers and 2 erasers or not, we need to substitute
[tex]x = 2 \: \: and \: \: y = 2[/tex]
in to the inequality.
If the two values satisfy the inequality, then it is possible.
[tex]0.35(2) + 0.99(2) + 0.59 \leqslant 4[/tex]
This gives us,
.
[tex]0.7 + 1.98+ 0.59 \leqslant 4[/tex]
[tex]3.27 \leqslant 4[/tex]
The above statement is true, therefore it is possible.
The correct answer is A.
