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enyata [817]
3 years ago
9

Expand the expression and combine like terms: 4(1.75y−3.5)+1.25y

Mathematics
1 answer:
Bas_tet [7]3 years ago
8 0
4(1.75y−3.5)+1.25y . First multiply through with 4 the numbers in the parentheses. 7y➖ 14 ➕ 1.25y add 1.25y to 7y That will be 8.25y So the answer will be 8.25y ➖ 14
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!Can some1 help! !Help FAST!
NemiM [27]

Answer:

4y-9

Step-by-step explanation:

y is the number, 4 times the number = 4y

so 4y-9

4 0
3 years ago
Read 2 more answers
Which shows how the distributive property can be used to evaluate 7x8 4/5?
Yuri [45]

Answer:

61 3/5

Step-by-step explanation:

we realize that 8 4/5 can be written as [8 + (4/5)]

hence 7 x 8 4/5

= 7 x [8 + (4/5)]

= 7 [8 + (4/5)]   (use the distributive property, see attached for reference)

= 7(8) + 7(4/5)

= 56 + 28/5  (convert 28/5 into mixed fraction)

= 56 + 5 3/5

= 61 3/5 (answer)

5 0
2 years ago
Read 2 more answers
A standard deck of cards has 52 cards, 4 of each type (Ace, King, Queen, Jack, 10,...,2). From a well-shuffled deck, you are dea
Wewaii [24]

Answer:

a)%85,5

b)%92,2

Step-by-step explanation:

a) To determine the probability of getting a hand with at least one face, we need to calculate the probability of the hand without any face firstly.

(36/52)*(35/51)*(34/50)*(33/49)*(32/48)=0,145

Then, we need to deduct this value from the probability 1

1-0,145=0,855

The probability of the hand with at least one face is %85,5.

b)To determine the probability of a hand with at least one ace and face we will track the same road again.

(32/52)*(31/51)*(30/50)*(29/49)*(28/48)=0,0775

Then, we need to deduct this value from the probability 1

1-0,0775=0,922

The probability of the hand with at least one ace and one face is %92,2.

3 0
2 years ago
A sneaker company can make 50 pairs of sneakers for every 6 yards of material. Make a table to show the relation between materia
Law Incorporation [45]

A sneaker company can make 50 pairs of sneakers for every 6 yards of material.

Then in each case you can consider proportion.

1. 50 pairs -- 6 yards,

80 pairs -- x yards.

Mathematically,

\dfrac{50}{80}=\dfrac{6}{x},\\ \\x=\dfrac{80\cdot 6}{50}=9.6\ yards.

2. 50 pairs -- 6 yards,

100 pairs -- x yards.

Mathematically,

\dfrac{50}{100}=\dfrac{6}{x},\\ \\x=\dfrac{100\cdot 6}{50}=12\ yards.

3. 50 pairs -- 6 yards,

120 pairs -- x yards.

Mathematically,

\dfrac{50}{120}=\dfrac{6}{x},\\ \\x=\dfrac{120\cdot 6}{50}=14.4\ yards.

4. 50 pairs -- 6 yards,

140 pairs -- x yards.

Mathematically,

\dfrac{50}{140}=\dfrac{6}{x},\\ \\x=\dfrac{140\cdot 6}{50}=16.8\ yards.

Thus, the table is

\begin{array}{cc}\text{Pairs}&\text{Yards}\\50&6\\80&9.6\\100&12\\120&14.4\\140&16.8\end{array}

The domain of the relation is the set \{50,80,100,120,140\} and the range of the relation is \{6,9.6,12,14.4,16.8\}.See attached diagram for the graph of the relation.

4 0
2 years ago
What is the answer?​
ella [17]
<h2><em>I believe 10.91ft but I'm not sure.</em></h2>
4 0
3 years ago
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