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Juliette [100K]
3 years ago
5

Can you help me please I pay you £50

Mathematics
1 answer:
trapecia [35]2 years ago
7 0

Answer:

(-3,4), (-3,1), and (-5,1)

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Tim's bean sprout grew 3/ inches. Teegan's bean sprout grew 2 3/4 inches. How many more inches did Tim's bean sprout grow than T
ki77a [65]

Answer:

Tim's bean sprout grow by more than \frac{5}{8} \text{ inches}.

Step-by-step explanation:

We are given that Tim's bean sprout grew 3 3/8 inches. Teegan's bean sprout grew 2 3/4 inches.

We have to find how many more inches did Tim's bean sprout grow than Teegan's.

Firstly, converting both the mixed fractions in improper fraction get;

Tim's bean sprout grew = 3\frac{3}{8} = \frac{27}{8} \text{ inches}

Teegan's bean sprout grew = 2\frac{3}{4} = \frac{11}{4} \text{ inches}

Since the denominator of both the fractions is not the same, so we can't compare them both as which is larger or smaller.

Tim's bean sprout grew = \frac{27\times 1}{8\times 1} = \frac{27}{8} \text{ inches}

Teegan's bean sprout grew = \frac{11 \times 2}{4\times 2}=\frac{22}{8} \text{ inches}

Now, we can clearly see that Tim's bean sprout grew more than Teegan's bean sprout.

So, Tim's bean sprout grow by more than = \frac{27}{8}- \frac{22}{8} = \frac{5}{8} \text{ inches}

Hence, Tim's bean sprout grow by more than \frac{5}{8} \text{ inches}.

8 0
3 years ago
Martin decided to buy 2 bags of grapes weighing 5 and two-thirds pounds each, instead of the 3 bags weighing 4 and one-fifth pou
Juli2301 [7.4K]

Answer: He ended up with about 1 pound less grapes

Step-by-step explanation:

2 bags of grapes weighing 5 and two-thirds pounds each will give a total weight of:

= 2 × 5 2/3

= 2 × 17/3

= 34/3

= 11 1/3 pounds

3 bags weighing 4 and one-fifth pounds each, will give a weight of:

= 3 × 4 1/5

= 3 × 21/5

= 63/5

= 12 3/5

He ended up buying lesser grapes. This was about 12 3/5 - 11 1/3 = 1 4/15 less. This means that he ended up with about 1 pound less grapes

3 0
3 years ago
Read 2 more answers
How many nineths does it take to make the same as 1/3
krek1111 [17]

Answer:

3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
I have some questions regarding combinations in finite. Here is my first one I am stumped on:
podryga [215]

We will get the number of possible selections, and then subtract the number less than 25 cents.

We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.

That's 5*4*3  = 60 selections 

Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels. 

To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2  on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).

If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.

And there is 1 other selection less than 25 cents, 2 dimes and no nickels. 

So that's 4+3+1 = 8 selections which we must subtract from the 60.
 
Answer 60-8 = 52 selections of coins worth 25 cents or more.

8 0
3 years ago
Slope from two points: (4, 4) and (7, -2)
BaLLatris [955]

Answer:

2

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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