890678340 in word form

Drag the tiles to match each measurement with an equivalent measurement.

KonstantinChe [14]

Answer:

Measurement Equivalent Measurement

6 pints 12 cups

16 pints 8 quarts

8 fluid ounces 1 cup

8 pints 1 gallon

Step-by-step explanation:

1 gallon = 4 quarts

1 quart = 2 pints

1 pint = 2 cups

1 cup = 8 fluid ounces

Measurement Equivalent Measurement

6 pints 12 cups

16 pints 8 quarts

8 fluid ounces 1 cup

8 pints 1 gallon

8 0

3 years ago

At a candy store you can pay $16.10 for 7 giant lollipop what would be the cost for 6 lollipops?

Alex Ar [27]

Answer: $13.80 for 6 giant lollipops

Step-by-step explanation: To find the cost of 6 giant lollipops at the candy store, we first need to find the unit price for a giant lollipop.

Unit price means the cost per unit,

in this case, the cost per giant lollipop.

Since we know that it costs $16.10 for 7 giant lollipops, to find the cost

for 1 giant lollipop, we need to 16.10 by 7.

16.10 divided by 7 is 2.3.

This means that each lollipop is $2.30.

Now to find the cost for 6, we multiply

the cost per 1 lollipop by 6.

So we have 2.30 × 6 which is $13.80.

So the cost for 6 lollipops at the candy store is $13.80.

8 0

3 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$