Jaya's coach agreed to buy ice

A shoe store is selling 3 pairs of shoes for $26.25 . If a mother is buying 4 pairs of shoes for her children for school, how mu

elena-s [515]

Answer: $35

Step-by-step explanation: If three pairs of shoes cost 26.25 you need to divide that by 3 to get the individual price of each pair of shoes, then add the price of the shoe individually to 26.25 and you will get your answer.

7 0

2 years ago

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Which property of multiplication is shown? 6 · 1 = 1 · 6

saw5 [17]

Answer: Commutative property of multiplication

Step-by-step explanation: The problem 6 · 1 = 1 · 6 demonstrates the commutative property of multiplication.

In other words, the commutative property of multiplication says that changing the order of the factors does not change the product.

So for example here, 6 · 1 is equal to 6 and 1 · 6 also equals 6.

Since 6 = 6, we can easily see that 6 · 1 must be equal to 1 · 6.

In more general terms, the commutative property of multiplication can be written as a · b = b · a where a and b are variables that can represent any numbers.

5 0

2 years ago

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Help me with 16,17,18,and 19 please simpl

mixas84 [53]

Answer:

Ques 16)

We have to simplify the expression:

$\dfrac{t^2}{t^2+3t-18}-(\dfrac{5t}{t^2+3t-18}-\dfrac{t-3}{t^2+3t-18})\\ \\=\dfrac{t^2}{t^2+3t-18}-(\dfrac{4t+3}{t^2+3t-18})\\ \\=\dfrac{t^2-4t-3}{t^2+3t+18}$

Ques 17)

$\dfrac{3w^2+7w-7}{w^2+8w+15}+\dfrac{2w^2-9w+4}{(2w^2+9w-5)(w^2-w-12)}\\ \\=\dfrac{3w^2+7w-7}{w^2+8w+15}+\dfrac{(2w-1)(w-4)}{(2w-1)(w+5)(w+3)(w-4)}\\\\=\dfrac{3w^2+7w-7}{(w+3)(w+5)}+\dfrac{1}{(w+3)(w+5)}\\\\=\dfrac{3w^2+7w-7+1}{(w+3)(w+5)}\\\\=\dfrac{(3w-2)(w+3)}{(w+5)(w+3)}\\\\=\dfrac{3w-2}{w+5}$

Ques 18)

Let the blank space be denoted by the quantity 'x'.

$\dfrac{x}{12a^2+8a}+\dfrac{15a^2}{12a^2+8a}=\dfrac{7a}{3a+2}\\ \\\dfrac{x+15a^2}{12a^2+8a}=\dfrac{7a}{3a+2}\\\\=\dfrac{x+15a^2}{4a(3a+2)}=\dfrac{7a}{3a+2}\\\\=\dfrac{x+15a^2}{4a}=7a\\\\x+15a^2=28a^2\\\\x=28a^2-15a^2\\\\x=13a^2$

Ques 19)

Let the missing quantity be denoted by 'x'.

$\dfrac{p^2+7p+2}{p^2+5p-14}-\dfrac{x}{p^2+5p-14}=\dfrac{p-1}{p-2}\\ \\\dfrac{p^2+7p+2-x}{p^2+5p-14}=\dfrac{p-1}{p-2}\\\\\dfrac{p^2+7p+2-x}{(p-2)(p+7)}=\dfrac{p-1}{p-2}\\\\p^2+7p+2-x=(p-1)(p+7)\\\\p^2+7p+2-x=p^2+6p-7\\\\x=p+9$