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Volgvan
3 years ago
14

o Marcella bought a 25-ounce bottle of olive oil for $5.88. She used 60% of the olive oil in two weeks. Which of the following i

s closest to the cost of the oil she used? F $0.24 G $0.39 H $2.35 J $3.53

Mathematics
1 answer:
Irina18 [472]3 years ago
7 0
First divide the price of the olive oil by the volume of the olive oil to find the price per ounce.  5.88/25oz=0.2352 dollars/ounce.  Then you have to multiply the volume of the bottle by .60 to find what 60% of the volume is. 0.60x25=15 ounces.  since in the first two weeks she used 15 ounces which has a price of 0.2352 dollars/ounce you have to multiply 15 by 0.2352 which equals 3.528 dollars.
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Add [1/-4 3/5] [-2/6 -2/4]
brilliants [131]

Answer:

​25/138

Step-by-step explanation:

1 Convert 4\frac{3}{5}4

​5

​

​3

​​  to improper fraction. Use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

​c

​

​b

​​ =

​c

​

​ac+b

​​ .

\frac{1}{-(\frac{4\times 5+3}{5})}(-\frac{2}{6}-\frac{2}{4})

​−(

​5

​

​4×5+3

​​ )

​

​1

​​ (−

​6

​

​2

​​ −

​4

​

​2

​​ )

2 Simplify  4\times 54×5  to  2020.

\frac{1}{-(\frac{20+3}{5})}(-\frac{2}{6}-\frac{2}{4})

​−(

​5

​

​20+3

​​ )

​

​1

​​ (−

​6

​

​2

​​ −

​4

​

​2

​​ )

3 Simplify  20+320+3  to  2323.

\frac{1}{-(\frac{23}{5})}(-\frac{2}{6}-\frac{2}{4})

​−(

​5

​

​23

​​ )

​

​1

​​ (−

​6

​

​2

​​ −

​4

​

​2

​​ )

4 Simplify  \frac{2}{6}

​6

​

​2

​​   to  \frac{1}{3}

​3

​

​1

​​ .

\frac{1}{-(\frac{23}{5})}(-\frac{1}{3}-\frac{2}{4})

​−(

​5

​

​23

​​ )

​

​1

​​ (−

​3

​

​1

​​ −

​4

​

​2

​​ )

5 Simplify  \frac{2}{4}

​4

​

​2

​​   to  \frac{1}{2}

​2

​

​1

​​ .

\frac{1}{-(\frac{23}{5})}(-\frac{1}{3}-\frac{1}{2})

​−(

​5

​

​23

​​ )

​

​1

​​ (−

​3

​

​1

​​ −

​2

​

​1

​​ )

6 Find the Least Common Denominator (LCD) of \frac{1}{3},\frac{1}{2}

​3

​

​1

​​ ,

​2

​

​1

​​ . In other words, find the Least Common Multiple (LCM) of 3,23,2.

LCD = 66

7 Make the denominators the same as the LCD.

-\frac{1\times 2}{3\times 2}-\frac{1\times 3}{2\times 3}−

​3×2

​

​1×2

​​ −

​2×3

​

​1×3

​​

8 Simplify. Denominators are now the same.

-\frac{2}{6}-\frac{3}{6}−

​6

​

​2

​​ −

​6

​

​3

​​

9 Join the denominators.

\frac{-2-3}{6}

​6

​

​−2−3

​​

10 Simplify  -\frac{1}{3}-\frac{1}{2}−

​3

​

​1

​​ −

​2

​

​1

​​   to  -\frac{5}{6}−

​6

​

​5

​​ .

\frac{1}{-(\frac{23}{5})}\times \frac{-5}{6}

​−(

​5

​

​23

​​ )

​

​1

​​ ×

​6

​

​−5

​​

11 Move the negative sign to the left.

-\frac{1}{\frac{23}{5}}\times \frac{-5}{6}−

​

​5

​

​23

​​

​

​1

​​ ×

​6

​

​−5

​​

12 Invert and multiply.

-\frac{5}{23}\times \frac{-5}{6}−

​23

​

​5

​​ ×

​6

​

​−5

​​

13 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}

​b

​

​a

​​ ×

​d

​

​c

​​ =

​bd

​

​ac

​​ .

-\frac{5\times -5}{23\times 6}−

​23×6

​

​5×−5

​​

14 Simplify  5\times -55×−5  to  -25−25.

-\frac{-25}{23\times 6}−

​23×6

​

​−25

​​

15 Simplify  23\times 623×6  to  138138.

-\frac{-25}{138}−

​138

​

​−25

​​

16 Move the negative sign to the left.

-(-\frac{25}{138})−(−

​138

​

​25

​​ )

17 Remove parentheses.

\frac{25}{138}

​138

​

​25

​​

7 0
3 years ago
A certain substance doubles its volume every minute at 9 AM a small amount is placed in a container ,at 10 AM the container was
Y_Kistochka [10]

Answer:

the container is 1/4 full at 9:58 AM

Step-by-step explanation:

since the volume doubles every minute , the formula for calculating the volume V at any time t is

V(t)=V₀*2^-t , where t is in minutes back from 10 AM and V₀= container volume

thus for t=1 min (9:59 AM) the volume is V₁=V₀/2 (half of the initial one) , for t=2 (9:58 AM) is V₂=V₁/2=V₀/4  ...

therefore when the container is 1/4 full the volume is V=V₀/4 , thus replacing in the equation we obtain

V=V₀*2^-t

V₀/4 = V₀*2^-t

1/4 = 2^-t

appling logarithms

ln (1/4) = -t* ln 2

t = - ln (1/4)/ln 2 = ln 4 /ln 2 = 2*ln 2 / ln 2 = 2

thus t=2 min before 10 AM → 9:58 AM

therefore the container is 1/4 full at 9:58 AM

8 0
3 years ago
Strontium-90 has a half-life of about 29 years. After 65 years, about how many grams of a 25-gram sample will remain?
konstantin123 [22]

Answer:

Ending Amount = Beginning Amount / 2 ^ n

where 'n' = number of half-lives n = 65 / 29 = 2.2413793103

Ending Amount = 25 g / 2^2.2413793103

Ending Amount = 25 / 4.7284892286

Ending Amount = 5.2871009727  g = 5.29 g (rounded)

Step-by-step explanation:

3 0
3 years ago
Plz answer below 38 pts and will mkae u brainliest.
mestny [16]

Answer:

Question 1: She needs 12 blue beads and 48 black beads.

Question 2: She needs 10 blue beads and 25 black beads.

Step-by-step explanation:

I used ratios

question one's ratio is 3:12 and the second one's ratio is 4:10. If you keep that ratio consistent than those are the numbers you get based on the number of beads long she wants her necklace.

that's my best guess

7 0
3 years ago
Can anybody please help
aliya0001 [1]
Y=20 X=10 make this brainliest ❤️
3 0
3 years ago
Read 2 more answers
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